Circuits

(D) For the 6 µF and 3 µF capacitor in series, the equivalent capacitance is 2 µF. Adding the 2 µF in parallel gives a total capacitance of 4 µF

Three capacitors are connected to a 5 V source, as shown in the circuit diagram above The equivalent capacitance for the circuit is (A) 1/11 µF (B) 11/18 µF (C) 1 µF (D) 4 µF (E) 11 µF

(B) The resistances are as follows: I: 2 Ω, II: 4 Ω, III: 1 Ω, IV: 2 Ω

Which two arrangements of resistors shown above have the same resistance between the terminals? (A) I and II (B) I and IV (C) II and III (D) II and IV (E) III and IV

(D) P = IV = 1.56 kW. Energy = Pt = 1.56 kW × 8 h = 12.48 kW-h

An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours? (A) $0.19 (B) $0.29 (C) $0.75 (D) $1.25 (E) $1.55

(E) R = ρL/A. Least resistance is the widest, shortest resistor

The five resistors shown below have the lengths and cross-sectional areas indicated and are made of material with the same resistivity. Which resistor has the least resistance?

(A) S1 must be closed to have any current. The greatest voltage will occur with the greatest current through R3 but closing S2 or S3 will draw current away from R3 ****use***

In the circuit above, the resistors all have the same resistance. The battery, wires, and ammeter have negligible resistance. A closed switch also has negligible resistance Closing which of the switches will produce the greatest voltage across R 3 ? (A) S 1 only (B) S 2 only (C) S 1 and S 2 only (D) S 1 and S 3 only (E) S 1 , S 2 , and S 3

(E) Most rapid heating requires the largest power dissipation. This occurs with the resistors in parallel.

Suppose you are given a constant voltage source V0 and three resistors R1, R2, and R3 with R1 > R2 > R3. If you wish to heat water in a pail which of the following combinations of resistors will give the most rapid heating?

(E) The equivalent resistance through path ACD is equal to the equivalent resistance through path ABD, making the current through the two branches equal

A 9-volt battery is connected to four resistors to form a simple circuit as shown above. How would the current through the 2 ohm resistor compare to the current through the 4 ohm resistor? (A) one-forth as large (B) one-half as large (C) four times as large (D) twice as large (E) equally as large

(B) P = V²/R and R = ρL/A giving P = V²A/ρL

A fixed voltage is applied across the length of a tungsten wire. An increase in the power dissipated by the wire would result if which of the following could be increased? (A) The resistivity of the tungsten (B) The cross-sectional area of the wire (C) The length of the wire (D) The temperature of the wire (E) The temperature of the wire's surroundings

(D) ε = IR_total where R_total = 35 Ω

An ideal battery, an ideal ammeter, a switch and three resistors are connected as shown. With the switch open as shown in the diagram the ammeter reads 2.0 amperes. With the switch open, what must be the voltage supplied by the battery? (A) 30 V (B) 40 V (C) 60 V (D) 70 V (E) 110 V

(A) V = IR

An ideal battery, an ideal ammeter, a switch and three resistors are connected as shown. With the switch open as shown in the diagram the ammeter reads 2.0 amperes. With the switch open, what would be the potential difference across the 15 ohm resistor? (A) 30 V (B) 40 V (C) 60 V (D) 70 V (E) 110V

(D) Shorting bulb 4 decreases the resistance in the right branch, increasing the current through bulb 3 and in the main branch containing bulb 1.

For the circuit shown, when a shorting wire (no resistance) connects the points labeled A and B, which of the numbered light bulbs become brighter? Assume that all four bulbs are identical and have resistance R . (A) Bulb 1 only (B) Bulb 2 only (C) Bulb 3 only (D) Bulbs 1 and 3 only (E) Bulbs 1, 2, and 3

(A) The resistances are, respectively, 4/3 R, 2/5 R, R, and 5/3 R

Given 4 identical resistors of resistance R, which of the following circuits would have an equivalent resistance of 4/3 R?

(A) R = V/I where V = W/Q and Q = It giving R = W/I²t and W = joules = kg m²/s²

In terms of the seven fundamental SI units in the MKS system, the Ohm is written as In terms of the seven fundamental SI units in the MKS system, the Ohm is written as (A) (kg*m²)/(A²*s³) (B) (kg*m²*s)/(C²) (C) (kg*m)/(C*s) (D) (kg*m²)/(A*s²) (E) (kg*s²)/(A²*m²)

(C) With a total resistance of 10 Ω, the total current is 1.2 A. The terminal voltage V_T = ε- Ir

In the circuit above the voltmeter V draws negligible current and the internal resistance of the battery is 1.0 ohm. The reading of the voltmeter is (A) 10.5 V (B) 12.0 V (C) 10.8 V (D) 13.0 V (E) 11.6 V

(B) Using ratios, the currents in the 6 Ω and 3 Ω resistors are 1 A and 2 A. They have three times and 3/2 times the resistance of the 2 Ω resistor so they will have 1/3 and 2/3 the current. The total current is then 6 A giving a potential drop of 36 V across the 6 Ω resistor in the main branch and adding any one of the branches below with the loop rule gives 36 V + 6 V = 42 V for the battery

In the electric circuit shown above, the current through the 2.0 Ω resistor is 3.0 A. Approximately what is the emf of the battery? (A) 51 V (B) 42 V (C) 36 V (D) 24 V (E) 21 V

(B) In parallel 1/R_t = ∑ 1/R

The equivalent resistance of the circuit shown to the right with resistances R1 = 4.00 Ω, R2 = 3.00 Ω, and R3 = 2.00 Ω is (A) 0.111 Ω (B) 0.923 Ω (C) 1.08 Ω (D) 3.00 Ω (E) 9.00 Ω

(D) R = ρL/A

Two cables can be used to wire a circuit. Cable A has a lower resistivity, a larger diameter, and a different length than cable B. Which cable should be used to minimize heat loss if the same current is maintained in either cable? (A) Cable A (B) Cable B (C) The heat loss is the same for both. (D) It cannot be determined without knowing the length of each cable. (E) It cannot be determined without knowing the materials contained in each cable

(A) Superconductors have a property where the resistance goes to zero below a certain threshold temperature

Which of the following graphs would best represent the resistance versus temperature relationship for a superconductor?

(C) R = ρL/A ∝ L/d2 where d is the diameter. R_II/R_I =L_II/d_II² ÷ L_I/d_I² = (2L_I)d_I²/[L_I(2d_I)2] = ½

Wire I and wire II are made of the same material. Wire II has twice the diameter and twice the length of wire I. If wire I has resistance R, wire II has resistance (A) R/8 (B) R/4 (C) R/2 (D) R (E) 2R

(C) Resistivity is dependent on the material. Not to be confused with resistance

A cylindrical graphite resistor has length L and cross-sectional area A. It is to be placed into a circuit, but it first must be cut in half so that the new length is ½ L. What is the ratio of the new resistivity to the old resistivity of the cylindrical resistor? (A) 4 (B) 2 (C) 1 (D) ½ (E) ¼

(A) 1 kW-h = 1000 W × 60 min = 60,000 W-min = I²Rt = I²(20 Ω)(30 min)

In a 30-minute interval, one kilowatt-hour of electrical energy is dissipated in a resistance of 20 ohms by a current of (A) 10 amp. (B) 20 amp. (C) 14.1 amp. (D) 36 amp. (E) 18 amp.

(A) Breaking the circuit in the lower branch lowers the total current in the circuit, decreasing the voltage across R1. Looking at the upper loop, this means R2 now has a larger share of the battery voltage and the voltage across AD is the same as the voltage across BC

The circuit shown has an ideal ammeter with zero resistance and four identical resistance light bulbs which are initially illuminated. A person removes the bulb R4 from its socket thereby permanently breaking the electrical circuit at that point. Which statement is true of the circuit after removing the bulb? (A) The voltage from B → C increases. (B) The power supplied by the battery increases (C) The voltage across R1 (D) The ammeter reading is unchanged. increases. (E) The bulb R2 maintains the same brightness

(D) The voltmeter is essentially another resistor. The voltmeter in parallel with the 100 Ω resistor acts as a 500 Ω resistor, which will half ½ the voltage of the 100 Ω resistor on the left. Thus the 120 V will split into 80 V for the 1000 Ω resistor and 40 V for the voltmeter combination.

Two 1000 Ω resistors are connected in series to a 120-volt electrical source. A voltmeter with a resistance of 1000 Ω is connected across the last resistor as shown. What would be the reading on the voltmeter? (A) 120 V (B) 80 V (C) 60 V (D) 40 V (E) 30 V

(A) Starting at A and summing potential differences counterclockwise to point C gives 12 V

An electric circuit consists of a 12 V battery, an ideal 10 A fuse, and three 2 Ω resistors connected as shown above What would be the reading on a voltmeter connected across points A and C ? (A) 12 V (B) 6 V (C) 3 V (D) 2 V (E) 0 V, since the fuse would break the circuit

(D) With B2 burning out, the total resistance of the circuit increases as it is now a series circuit. This decreases the current in the main branch, decreasing V1. For V1 to be halved, the current must be halved which means the total resistance must be doubled, which by inspection did not happen in this case (total before = 5/3 R, total after = 3 R)

B1, B2, B3, and B4 are identical light bulbs. There are six voltmeters connected to the circuit as shown. All voltmeters are connected so that they display positive voltages. Assume that the voltmeters do not affect the circuit. If B 2 were to burn out, opening the circuit, what would happen to the reading of V 1 ? Let V be its original reading when all bulbs are functioning and let V be its reading when B 2 is burnt out. (A) V > 2V (B) 2V > V > V (C) V = V (D) V > V > V/2 (E) V/2 > V

(B) Even though B2 burns out, the circuit is still operating elsewhere as there are still closed paths.

B1, B2, B3, and B4 are identical light bulbs. There are six voltmeters connected to the circuit as shown. All voltmeters are connected so that they display positive voltages. Assume that the voltmeters do not affect the circuit. If B 2 were to burn out, opening the circuit, which voltmeter(s) would read zero volts? (A) none would read zero (B) only V 2 (C) only V 3 and V 4 (D) only V 2 , V 4 , and V 5 (E) they would all read zero

(C) If A were to burn out, the total resistance of the parallel part of the circuit increases, causing less current from the battery and less current through bulb A. However, A and B split the voltage from the battery in a loop and with less current through bulb A, A will have a smaller share of voltage, increasing the potential difference (and the current) through bulb B.

Consider a simple circuit containing a battery and three light bulbs. Bulb A is wired in parallel with bulb B and this combination is wired in series with bulb C. What would happen to the brightness of the other two bulbs if bulb A were to burn out? (A) There would be no change in the brightness of either bulb B or bulb C. (B) Both would get brighter. (C) Bulb B would get brighter and bulb C would get dimmer. (D) Bulb B would get dimmer and bulb C would get brighter. (E) Only bulb B would get brighter

(D) When the current is 0.5 A, the voltage across the resistor is V = IR = 5 V. According to the loop rule, the remaining 7 V must be across the capacitor.

For the RC circuit shown, the resistance is R = 10.0 Ω, the capacitance is C = 5.0 F and the battery has voltage ξ = 12 volts . The capacitor is initially uncharged when the switch S is closed at time t = 0. At some time later, the current in the circuit is 0.50 A. What is the magnitude of the voltage across the capacitor at that moment? (A) 0 volts (B) 5 volts (C) 6 volts (D) 7 volts (E) 12 volts

(E) The equivalent resistance of the two 4 Ω resistors on the right is 2 Ω making the total circuit resistance 10 Ω and the total current 2.4 A. The 2.4 A will divide equally between the two branches on the right. Q = It = (1.2 A)(5 s) = 6 C

How many coulombs will pass through the identified resistor in 5 seconds once the circuit was closed? (A) 1.2 (B) 12 (C) 2.4 (D) 24 (E) 6

(A) The greatest current is in the main branch

The diagram above shows five resistors connected to a voltage source. Which resistor has the greatest electric current through it? (A) 1 Ω (B) 2 Ω (C) 3 Ω (D) 4 Ω (E) 5 Ω

(D) Resistors J and N are in the main branch and therefore receive the largest current

The diagram below shows five identical resistors connected in a combination series and parallel circuit to a voltage source. Through which resistor(s) would there be the greatest current? (A) J only (B) M only (C) N only (D) J&N only (E) K&L only

(C) Closing the switch adds another parallel branch, increasing the total current delivered by the battery. Bulb 3 will get brighter. Bulb 2, in its own loop with bulb 3 and the battery will then lose some of its share of the potential difference from the battery and will get dimmer.

The three lightbulbs in the circuit above are identical, and the battery has zero internal resistance. When switch S is closed to cause bulb 1 to light, which of the other two bulbs increase(s) in brightness? (A) Neither bulb (B) Bulb 2 only (C) Bulb 3 only (D) Both bulbs (E) It cannot be determined without knowing the emf of the battery.

(C) In series the capacitors have the same charge, but the smaller capacitor will have the larger potential difference (to force the same charge on a smaller area)

Three capacitors are connected to a 5 V source, as shown in the circuit diagram above How do the charge Q3 stored in the 3 µF capacitor and the voltage V 3 across it compare with those of the 6 µF capacitor? Charge Voltage (A) Q3 < Q6 V 3 = V 6 (B) Q3 = Q6 V 3 < V 6 (C) Q3 = Q6 V 3 > V 6 (D) Q3 > Q6 V 3 = V 6 (E) Q3 > Q6 V 3 > V 6

(E) In a simple series circuit with two batteries opposing one another the voltages subtract from one another. The total effective voltage for this circuit is then 4 V. With a total resistance of 20 Ω the total current is (4 V)/(20 Ω)

When the switch is closed, what would be the current in the circuit shown in the diagram above if the two batteries are opposing one another? (A) 1.25 A (B) 0.75 A (C) 0.5 A (D) 0.3 A (E) 0.2 A

(B) Voltmeters must be placed in parallel and ammeters must be placed in series.

Which of the following wiring diagrams could be used to experimentally determine R using Ohm's Law? Assume an ideal voltmeter and an ideal ammeter.

(A) P = IV

A heating coil is rated 1200 watts and 120 volts. What is the maximum value of the current under these conditions? (A) 10.0 A (B) 12.0 A (C) 14.1 A (D) 0.100 A (E) 0.141 A

(D) P=I²R

A wire of resistance R dissipates power P when a current I passes through it. The wire is replaced by another wire with resistance 3R. The power dissipated by the new wire when the same current passes through it is (A) P/9 (B) P/3 (C) P (D) 3P (E) 6P

(B) W = Pt = I²Rt

An immersion heater of resistance R converts electrical energy into thermal energy that is transferred to the liquid in which the heater is immersed. If the current in the heater is I, the thermal energy transferred to the liquid in time t is (A) IRt (B) I²Rt (C) IR²t (D) IRt² (E) IR/t

(C) For each capacitor to have 6 µC, each branch will have 6 µC since the two capacitors in series in each branch has the same charge. The total charge for the three branches is then 18 µC. Q = CV gives 18 µC = (3 µF)V

Below is a system of six 2-microfarad capacitors. What potential difference must be applied between points X and Y so that the charge on each plate of each capacitor will have magnitude 6 microcoulombs? (A) 1.5 V (B) 3V (C) 6 V (D) 9 V (E) 18 V

(E) The current through bulb 3 is twice the current through 1 and 2 since the branch with bulb 3 is half the resistance of the upper branch. The potential difference is the same across each branch, but bulbs 1 and 2 must divide the potential difference between them.

Consider the compound circuit shown above. The three bulbs 1, 2, and 3 - represented as resistors in the diagram - are identical. Which of the following statements are true? I. Bulb 3 is brighter than bulb 1 or 2. II. Bulb 3 has more current passing through it than bulb 1 or 2. III. Bulb 3 has a greater voltage drop across it than bulb 1 or 2. (A) I only (B) II only (C) I & II only (D) I & III only (E) I, II, & III

(D) Each computer draws I = P/V = 4.17 A. 4 computers will draw 16.7 A, while 5 will draw over 20 A.

Each member of a family of six owns a computer rated at 500 watts in a 120 V circuit. If all computers are plugged into a single circuit protected by a 20 ampere fuse, what is the maximum number of the computers can be operating at the same time? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 or more

(E) If K burns out, the circuit becomes a series circuit with the three resistors, N, M and L all in series, reducing the current through bulb N.

Four identical light bulbs K, L, M, and N are connected in the electrical circuit shown above. Bulb K burns out. Which of the following statements is true? (A) All the light bulbs go out. (B) Only bulb N goes out. (C) Bulb N becomes brighter. (D) The brightness of bulb N remains the same. (E) Bulb N becomes dimmer but does not go out.

(A) Utilizing Kirchhoff's loop rule starting at the upper left and moving clockwise: - (2 A)(0.3 Ω) + 12 V - 6 V - (2 A)(0.2 Ω) -(2A)(R) - (2A)(1.5 Ω) = 0

In the circuit above, the emf's and the resistances have the values shown. The current I in the circuit is 2 amperes. The resistance R is (A) 1 Ω (B) 2Ω (C) 3 Ω (D) 4 Ω (E) 6 Ω

(B) The current through R is found using the junction rule at the top junction, where 1 A + 2 A enter giving I = 3 A. Now utilize Kirchhoff's loop rule through the left or right loops: (left side) + 16 V - (1 A)(4 Ω) - (3 A)R = 0 giving R = 4 Ω

In the circuit shown above, what is the resistance R? (A) 3 Ω (B) 4 Ω (C) 6 Ω (D) 12 Ω (E) 18 Ω

(C) When the capacitor is fully charged, the branch on the right has no current, effectively making the circuit a series circuit with the 100 Ω and 300 Ω resistors. Rtotal= 400 Ω, E = 10 V = IR

See the accompanying figure. What is the current through the 300 Ω resistor when the capacitor is fully charged? (A) zero (B) 0.020 A (C) 0.025 A (D) 0.033 A (E) 0.100 A

(D) Total circuit resistance (including internal resistance) = 40 Ω; total current = 0.3 A. ε = IR

The above circuit diagram shows a battery with an internal resistance of 4.0 ohms connected to a 16-ohm and a 20-ohm resistor in series. The current in the 20-ohm resistor is 0.3 amperes What is the emf of the battery? (A) 1.2 V (B) 6.0 V (C) 10.8 V (D) 12.0 V (E) 13.2 V

(A) The resistance of the two 2 Ω resistors in parallel is 1 Ω. Added to the 2 Ω resistor in series with the pair gives 3 Ω

The total equivalent resistance between points X and Y in the circuit shown above is (A) 3 Ω (B) 4 Ω (C) 5 Ω (D) 6 Ω (E) 7 Ω

(D) Using Kirchhoff's loop rule around the circuit going through either V or R since they are in parallel and will have the same potential drop gives: - V - (1.00 mA)(25 Ω) + 5.00 V - (1.00mA)(975 Ω) = 0

The voltmeter in the accompanying circuit diagram has internal resistance 10.0 kΩ and the ammeter has internal resistance 25.0 Ω. The ammeter reading is 1.00 mA. The voltmeter reading is most nearly: (A) 1.0 V (B) 2.0 V (C) 3.0 V (D) 4.0 V (E) 5.0 V

(B) R = ρL/A. Greatest resistance is the longest, narrowest resistor.

The five resistors shown below have the lengths and cross-sectional areas indicated and are made of material with the same resistivity. Which has the greatest resistance?

(D) In parallel, all the resistors have the same voltage (2 V). P3 = I3V3

Three different resistors R1, R2 and R3 are connected in parallel to a battery. Suppose R1 has 2 V across it, R2 = 4 Ω, and R3 dissipates 6 W. What is the current in R3? (A) 0.33 A (B) 0.5 A (C) 2 A (D) 3 A (E) 12 A

(B) The capacitance of the two capacitors in parallel is 2C. Combined with a capacitor in series gives C = (C*2C)/(C+2C) = 2/3C

Three identical capacitors each with a capacitance of C are connected as shown in the following diagram. What would be the total equivalent capacitance of the circuit? (A) 0.33 C (B) 0.67 C (C) 1.0 C (D) 1.5 C (E) 3.0 C

(D) In parallel V1 = V2. Q1 = C1V1 and Q2 = C2V2 so Q1/Q2 = C1/C2 = 1.5

Two capacitors are connected in parallel as shown above. A voltage V is applied to the pair. What is the ratio of charge stored on C₁ to the charge stored on C₂, when C₁ = 1.5C₂ ? (A) 4/9 (B) 2/3 (C) 1 (D) 3/2 (E) 9/4

(D) The larger loop, with twice the radius, has twice the circumference (length) and R = ρL/A

Two concentric circular loops of radii b and 2b, made of the same type of wire, lie in the plane of the page, as shown above. The total resistance of the wire loop of radius b is R. What is the resistance of the wire loop of radius 2b? (A) R/4 (B) R/2 (C) R (D) 2R (E) 4R

(B) Closing the switch reduces the resistance in the right side from 20 Ω to 15 Ω, making the total circuit resistance decrease from 35 Ω to 30 Ω, a slight decrease, causing a slight increase in current. For the current to double, the total resistance must be cut in half.

When the switch S is open in the circuit shown above, the reading on the ammeter A is 2.0 A. When the switch is closed, the reading on the ammeter is (A) doubled (B) increased slightly but not doubled (C) the same (D) decreased slightly but not halved (E) halved

(A) If the resistances are equal, they will all draw the same current

Which of the following statements is NOT true concerning the simple circuit shown where resistors R1, R2 and R3 all have equal resistances? (A) the largest current will pass through R1 (B) the voltage across R2 is 5 volts (C) the power dissipated in R3 could be 10 watts (D) if R2 were to burn out, current would still flow through both R1 and R3 (E) the net resistance of the circuit is less than R1

(D) Resistor D is in a branch by itself while resistors A, B and C are in series, drawing less current than resistor D.

If all of the resistors in the above simple circuit have the same resistance, which would dissipate the greatest power? (A) resistor A (B) resistor B (C) resistor C (D) resistor D (E) they would all dissipate the same power

(C) Summing the potential differences: - 6 V - (2 A)(0.2 Ω) - (2A)(1 Ω) = - 8.4 V

In the circuit above, the emf's and the resistances have the values shown. The current I in the circuit is 2 amperes. The potential difference between points X and Y is (A) 1.2 V (B) 6.0 V (C) 8.4 V (D) 10.8 V (E) 12.2 V

(E) Since the volume of material drawn into a new shape in unchanged, when the length is doubled, the area is halved. R = ρL/A

A cylindrical resistor has length L and radius r. This piece of material is then drawn so that it is a cylinder with new length 2L. What happens to the resistance of this material because of this process? (A) the resistance is quartered. (B) the resistance is halved. (C) the resistance is unchanged. (D) the resistance is doubled. (E) the resistance is quadrupled.

(C) P = V²/R

A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is (A) 0.1 Ω (B) 10 Ω (C) 12 Ω (D) 120 Ω (E) 1440 Ω

(D) P = IV

A household iron used to press clothes is marked "120 volt, 600 watt." In normal use, the current in it is (A) 0.2 A (B) 2 A (C) 4 A (D) 5 A (E) 7.2 A

(C) 1 year = 365 days × 24 hours/day = 8760 hours. W (energy) = Pt = 0.1 kW × 8760 hours = 867 kW-h × $0.10 per kW-h = $ 86.7

Approximately how much would it cost to keep a 100 W light bulb lit continuously for 1 year at a rate of $0.10 per kW ⋅ hr? (A) $1 (B) $10 (C) $100 (D) $1000 (E) $100000

(C) Wire CD shorts out bulb #3 so it will never light. Closing the switch merely adds bulb #2 in parallel to bulb #1, which does not change the potential difference across bulb #1

For the circuit shown, the ammeter reading is initially I. The switch in the circuit then is closed. Consequently: (A) The ammeter reading decreases. (B) The potential difference between E and F increases. (C) The potential difference between E and F stays the same. (D) Bulb #3 lights up more brightly. (E) The power supplied by the battery decreases.

(B) When the switch is closed, the circuit behaves as if the capacitor were just a wire, shorting out the resistor on the right.

In the circuit shown above, the battery supplies a constant voltage V when the switch S is closed. The value of the capacitance is C, and the value of the resistances are R1 and R2. Immediately after the switch is closed, the current supplied by the battery is (A) V/(R1 + R2) (B) V/R1 (C) V/R2 (D) V(R1 + R2)/R1R2 (E) zero

(E) To dissipate 24 W means R = V²/P = 6 Ω. The resistances, in order, are: 8 Ω, 4/3 Ω, 8/3 Ω, 12 Ω and 6 Ω

Which of the following combinations of 4Ω resistors would dissipate 24 W when connected to a 12 Volt battery?

(C) Resistance varies directly with temperature. Superconductors have a resistance that quickly goes to zero once the temperature lowers beyond a certain threshold.

Which of the following will cause the electrical resistance of certain materials known as superconductors to suddenly decrease to essentially zero? (A) Increasing the voltage applied to the material beyond a certain threshold voltage (B) Increasing the pressure applied to the material beyond a certain threshold pressure (C) Cooling the material below a certain threshold temperature (D) Stretching the material to a wire of sufficiently small diameter (E) Placing the material in a sufficiently large magnetic field

(C) Total resistance = E/I = 25 Ω. Resistance of the 30 Ω and 60 Ω resistors in parallel = 20 Ω adding the internal resistance in series with the external circuit gives Rtotal= 20 Ω + r = 25 Ω

A 30-ohm resistor and a 60-ohm resistor are connected as shown above to a battery of emf 20 volts and internal resistance r. The current in the circuit is 0.8 ampere. What is the value of r? (A) 0.22 Ω (B) 4.5 Ω (C) 5 Ω (D) 16Ω (E) 70 Ω

(D) Total circuit resistance of the load = R/2. Total circuit resistance including the internal resistance = r + R/2. The current is then E/(r + R/2) and the total power dissipated in the load is P = I2R_load= (ε²R/2)/(r + R/2)². Using calculus max/min methods or plotting this on a graph gives the value of R for which this equation is maximized of R = 2r. This max/min problem is not part of the B curriculum but you should be able to set up the equation to be maximized

A battery having emf E and internal resistance r is connected to a load consisting of two parallel resistors each having resistance R. At what value of R will the power dissipated in the load be a maximum? (A) 0 (B) r/2 (C) r (D) 2r (E) 4r

(D) Power = IV = 480 W = 0.48 kW. Energy = Pt = (0.48 kW)(2 hours) = 0.96 kWh

A certain coffeepot draws 4.0 A of current when it is operated on 120 V household lines. If electrical energy costs 10 cents per kilowatt-hour, how much does it cost to operate the coffeepot for 2 hours? (A) 2.4 cents (B) 4.8 cents (C) 8.0 cents (D) 9.6 cents (E) 16 cents

(A) Adding resistors in parallel decreases the total circuit resistance, this increasing the total current in the circuit.

A lamp, a voltmeter V, an ammeter A, and a battery with zero internal resistance are connected as shown above. Connecting another lamp in parallel with the first lamp as shown by the dashed lines would (A) increase the ammeter reading (B) decrease the ammeter reading (C) increase the voltmeter reading (D) decrease the voltmeter reading (E) produce no change in either meter reading

(D) Dimensional analysis: 1.6 × 10^-3 A = 1.6 × 10^-3 C/s ÷ 1.6 × 10^-19 C/proton 10^16 protons/sec ÷10^9 protons/meter = 10^7 m/s

A narrow beam of protons produces a current of 1.6 × 10⁻³ A. There are 10⁹ protons in each meter along the beam. Of the following, which is the best estimate of the average speed of the protons in the beam? (A) 10⁻¹⁵ m/s (B) 10⁻¹² m/s (C) 10⁻⁷ m/s (D) 10⁷ m/s (E) 10¹² m/s

(B) Kirchhoff's loop rule (V = Q/C for a capacitor)

A resistor R and a capacitor C are connected in series to a battery of terminal voltage V0. Which of the following equations relating the current I in the circuit and the charge Q on the capacitor describes this circuit? (A) V₀²+QC-I²R=0 (B) V₀² - Q/C - IR = 0 (C) V₀²-Q²/2C-I²R=0 (D) V₀-CI-I²R=0 (E) Q/C-IR=0

(A) P = V2/R and if V is constant P ∝ 1/R

A variable resistor is connected across a constant voltage source. Which of the following graphs represents the power P dissipated by the resistor as a function of its resistance R?

(B) R = ρL/A. If L ÷ 2, R ÷ 2 and is r ÷ 2 then A ÷ 4 and R × 4 making the net effect R ÷ 2 × 4

A wire of length L and radius r has a resistance R. What is the resistance of a second wire made from the same material that has a length L/2 and a radius r/2? (A) 4R (B) 2R (C) R (D) R/2 (E) R/4

(B) Resistance of bulbs B & C = 20 Ω combined with D in parallel gives 6.7 Ω for the right side. Combined with A & E in series gives a total resistance of 26.7 Ω. E = IR

Five identical light bulbs, each with a resistance of 10 ohms, are connected in a simple electrical circuit with a switch and a 10 volt battery as shown in the diagram below. The steady current in the above circuit would be closest to which of the following values? (A) 0.2 amp (B) 0.37 amp (C) 0.5 amp (D) 2.0 amp (E) 5.0 amp

(E) If M burns out, the circuit becomes a series circuit with the two resistors, N and K in series, with bulb L going out as well since it is in series with bulb M.

Four identical light bulbs K, L, M, and N are connected in the electrical circuit shown above. Bulb M burns out. Which of the following statements is true? (A) All the light bulbs go out. (B) Only bulb M goes out. (C) Bulb N goes out but at least one other bulb remains lit. (D) The brightness of bulb N remains the same. (E) Bulb N becomes dimmer but does not go out.

(B) The loop rule involves the potential and energy supplied by the battery and it's use around a circuit loop.

Kirchhoff's loop rule for circuit analysis is an expression of which of the following? (A) Conservation of charge (B) Conservation of energy (C) Ampere's law (D) Faraday's law (E) Ohm's law

(C) V_XY = ε - Ir where r is the internal resistance

The above circuit diagram shows a battery with an internal resistance of 4.0 ohms connected to a 16-ohm and a 20-ohm resistor in series. The current in the 20-ohm resistor is 0.3 amperes What is the potential difference across the terminals X and Y of the battery? (A) 1.2 V (B) 6.0 V (C) 10.8 V (D) 12.0 V (E) 13.2 V

(A) P = I²r

The above circuit diagram shows a battery with an internal resistance of 4.0 ohms connected to a 16-ohm and a 20-ohm resistor in series. The current in the 20-ohm resistor is 0.3 amperes What power is dissipated by the 4-ohm internal resistance of the battery? (A) 0.36 W (B) 1.2 W (C) 3.2 W (D) 3.6 W (E) 4.8 W

(E) Least power is for the greatest resistance (P = ε²/R)

The batteries in each of the circuits shown above are identical and the wires have negligible resistance. Which circuit dissipates the least power? (A) A (B) B (C) C (D) D (E) E

(A) Resistance of the 1 Ω and 3 Ω in series = 4 Ω. This, in parallel with the 2 Ω resistor gives (2 × 4) /(2 + 4) = 8/6 Ω. Also notice the equivalent resistance must be less than 2 Ω (the 2 Ω resistor is in parallel and the total resistance in parallel is smaller than the smallest resistor) and there is only one choice smaller than 2 Ω

The electrical resistance of the part of the circuit shown between point X and point Y is (A) 4/3 Ω (B) 2 Ω (C) 2.75 Ω (D) 4 Ω (E) 6 Ω

(C) V_T = ε - Ir

The emf of a battery is 12 volts. When the battery delivers a current of 0.5 ampere to a load, the potential difference between the terminals of the battery is 10 volts. The internal resistance of the battery is (A) 1 Ω (B) 2 Ω (C) 4 Ω (D) 20 Ω (E) 24 Ω

(D) The capacitance of the 4 µF and 2µF in parallel is 6 µF. Combined with the 3µF in series gives 2 µF for the right branch. Added to the 5 µF in parallel gives a total of 7 µF

The equivalent capacitance for this network is most nearly (A) 10/7 µF (B) 3/2 µF (C) 7/3 µF (D) 7 µF (E) 14 µF

(C) P = I²R and R = ρL/A giving P ∝ ρL/d²

The power dissipated in a wire carrying a constant electric current I may be written as a function of the length l of the wire, the diameter d of the wire, and the resistivity ρ of the material in the wire. In this expression, the power dissipated is directly proportional to which of the following? (A) l only (B) d only (C) l and ρ only (D) d and ρ only (E) l, d, and ρ

(A) By process of elimination, A is the only possible true statement

The total capacitance of several capacitors in parallel is the sum of the individual capacitance's for which of the following reasons? (A) The charge on each capacitor depends on its capacitance, but the potential difference across each is the same. (B) The charge is the same on each capacitor, but the potential difference across each capacitor depends on its capacitance. (C) Equivalent capacitance is always greater than the largest capacitance. (D) Capacitors in a circuit always combine like resistors in series. (E) The parallel combination increases the effective separation of the plates.

(E) In series 1/C_T = ∑1/C

Three 1/2 μF capacitors are connected in series as shown in the diagram above. The capacitance of the combination is (A) 0.1 μF (B) 1 μF (C) 2/3 μF (D) ½ μF (E) 1/6 μF

(C) There are several ways to do this problem. We can find the total energy stored and divide it into the three capacitors: U_C = ½ CV² = ½ (2 µF)(6 V)²= 36 µJ ÷ 3 = 12 µJ each

Three 6-microfarad capacitors are connected in series with a 6-volt battery. The energy stored in each capacitor is (A) 4 µJ (B) 6 µJ (C) 12 µJ (D) 18 µJ (E) 36 µJ

(B) In series 1/C_T = ∑1/C

Three 6-microfarad capacitors are connected in series with a 6-volt battery. The equivalent capacitance of the set of capacitors is (A) 0.5 µF (B) 2 µF (C) 3 µF (D) 9 µF (E) 18 µF

(B) The equivalent capacitance of the two 3 µF capacitors in parallel is 6 µF, combined with the 3 µF in series gives Ctotal= 2 µF

Three identical capacitors, each of capacitance 3.0 µF, are connected in a circuit with a 12 V battery as shown above. The equivalent capacitance between points X and Z is (A) 1.0 µF (B) 2.0 µF (C) 4.5 µF (D) 6.0 µF (E) 9.0 µF

(C) In series, they all have the same current, 2 A. P3 = I3V3

Three resistors - R1, R2, and R3 - are connected in series to a battery. Suppose R1 carries a current of 2.0 A, R2 has a resistance of 3.0 Ω, and R3 dissipates 6.0 W of power. What is the voltage across R3? (A) 1.0 V (B) 2.0 V (C) 3.0 V (D) 6.0 V (E) 12 V

(B) R = ρL/A ∝ L/d2 where d is the diameter. Rx/Ry = Lx/dx² ÷ Ly/dy²= (2Ly)dy²/[Ly(2dy)]² = ½

Two conducting cylindrical wires are made out of the same material. Wire X has twice the length and twice the diameter of wire Y. What is the ratio Rx/Ry (A) 1/4 (B) ½ (C) 1 (D) 2 (E) 4

(D) P = E²/R. Total resistance of n resistors in series is nR making the power P = E²/nR = P/n

When a single resistor is connected to a battery, a total power P is dissipated in the circuit. How much total power is dissipated in a circuit if n identical resistors are connected in series using the same battery? Assume the internal resistance of the battery is zero. (A) n²P (B) nP (C) P (D) P/n (E) P/n²

(C) The upper branch, with twice the resistance of the lower branch, will have ½ the current of the lower branch.

When there is a steady current in the circuit, the amount of charge passing a point per unit of time is (A) the same everywhere in the circuit (B) greater in the 1 Ω resistor than in the 2 Ω resistor (C) greater in the 2 Ω resistor than in the 3 Ω resistor (D) greater at point X than at point Y (E) greater in the 1 Ω resistor than in the 3 Ω resistor

(D) In series, the equivalent capacitance is calculated using reciprocals, like resistors in parallel. This results in an equivalent capacitance smaller than the smallest capacitor.

When two identical parallel-plate capacitors are connected in series, which of the following is true of the equivalent capacitance? (A) It depends on the charge on each capacitor. (B) It depends on the potential difference across both capacitors. (C) It is larger than the capacitance of each capacitor. (D) It is smaller than the capacitance of each capacitor. (E) It is the same as the capacitance of each capacitor.

(A) The equivalent resistance in parallel is smaller than the smallest resistance

When two resistors, having resistance R1 and R2, are connected in parallel, the equivalent resistance of the combination is 5 Ω. Which of the following statements about the resistances is correct? (A) Both R1 and R2, are greater than 5Ω (B) Both R1 and R2 are equal to 5 Ω (C) Both R1 and R2 are less than 5 Ω (D) The sum of R1 and R2 is 5 Ω (E) One of the resistances is greater than 5 Ω, one of the resistances is less than 5 Ω.

(A) R ∝ L/A = L/d². If d × 2, R ÷ 4 and if L ÷ 2, R ÷ 2 making the net effect R ÷ 8

Wire Y is made of the same material but has twice the diameter and half the length of wire X. If wire X has a resistance of R then wire Y would have a resistance of (A) R/8 (B) R/2 (C) R (D) 2R (E) 8R

(C) Using all three in series = 3 Ω, all three in parallel = 1/3 Ω. One in parallel with two in series = 2/3 Ω, one in series with two in parallel = 3/2 Ω

You are given three 1.0 Ω resistors. Which of the following equivalent resistances CANNOT be produced using all three resistors? (A) 1/3 Ω (B) 2/3 Ω (C) 1.0 Ω (D) 1.5 Ω (E) 3.0 Ω

(B) Closing the switch short circuits Bulb 2 causing no current to flow to it. Since the bulbs were originally in series, this decreases the total resistance and increases the total current, making bulb 1 brighter

The circuit in the figure above contains two identical lightbulbs in series with a battery. At first both bulbs glow with equal brightness. When switch S is closed, which of the following occurs to the bulbs? Bulb I | Bulb 2 (A) Goes out | Gets brighter (B) Gets brighter | Goes out (C) Gets brighter | Gets slightly dimmer (D) Gets slightly dimmer | Gets brighter (E) Nothing | Goes out

(C) P = Iε

The circuit shown above left is made up of a variable resistor and a battery with negligible internal resistance. A graph of the power P dissipated in the resistor as a function of the current I supplied by the battery is given above right. What is the emf of the battery? (A) 0.025 V (B) 0.67 V (C) 2.5 V (D) 6.25 V (E) 40 V

(D) Bulbs in the main branch have the most current through them and are the brightest.

The diagram above represents a simple electric circuit composed of 5 identical light bulbs and 2 flashlight cells. Which bulb (or bulbs) would you expect to be the brightest? (A) V only (B) V and W only (C) V and Z only (D) V, W and Z only (E) all five bulbs are the same brightness

(C) Summing the potential differences from bottom to top: left circuit: - (1 A)r + E = 10 V right circuit: + (1 A)r + E = 20 V, solve simultaneous equations

The figures above show parts of two circuits, each containing a battery of emf ε and internal resistance r. The current in each battery is 1 A, but the direction of the current in one battery is opposite to that in the other. If the potential differences across the batteries' terminals are 10 V and 20 V as shown, what are the values of ε and r ? (A) E = 5 V, r = 15 Ω (B) E =10 V, r = 100 Ω (C) E = 15 V, r = 5 Ω (D) E = 20 V, r = 10 Ω (E) The values cannot be computed unless the complete circuits are shown.

(D) For steady power dissipation, the circuit must allow current to slow indefinitely. For the greatest power, the total resistance should be the smallest value. These criteria are met with the resistors in parallel.

The five incomplete circuits below are composed of resistors R, all of equal resistance, and capacitors C, all of equal capacitance. A battery that can be used to complete any of the circuits is available. Into which circuit should the battery be connected to obtain the greatest steady power dissipation? (A) A (B) B (C) C (D) D (E) E

(C) Amperes = I (current); Volts = V (potential difference); Seconds = t (time): IVt = energy

The product (2 amperes × 2 volts × 2 seconds) is equal to (A) 8 coulombs (B) 8 newtons (C) 8 joules (D) 8 calories (E) 8 newton-amperes

(D) The equivalent capacitance between X and Y is twice the capacitance between Y and Z. Thismeans the voltage between X and Y is ½ the voltage between Y and Z. For a total of 12 V, this gives 4 V between X and Y and 8 V between Y and Z.

Three identical capacitors, each of capacitance 3.0 µF, are connected in a circuit with a 12 V battery as shown above. The potential difference between points Y and Z is (A) zero (B) 3 V (C) 4 V (D) 8 V (E) 9 V

(E) Summing the potential differences from left to right gives V_T = -12 V - (2 A)(2 Ω) = - 16 V. It is possible for V_T> E

A 12-volt storage battery, with an internal resistance of 2Ω, is being charged by a current of 2 amperes as shown in the diagram above. Under these circumstances, a voltmeter connected across the terminals of the battery will read (A) 4 V (B) 8 V (C) 10 V (D) 12 V (E) 16 V

(D) The resistance in each of the two paths is 9 Ω, making the current in each branch 1 A. From point A, the potential drop across the 7 Ω resistor is then 7 V and across the 4 Ω resistor is 4 V, making point B 3 V lower than point C

A 9-volt battery is connected to four resistors to form a simple circuit as shown above. What would be the potential at point B with respect to point C in the above circuit? (A) +7 V (B) +3 V (C) 0 V (D) -3 V (E) -7 V

(A) ACD = 9 Ω, ABD = 9 Ω so the total resistance is 4.5 Ω making the total current E/R = 2 A

A 9-volt battery is connected to four resistors to form a simple circuit as shown below. What would be the current at point E in the circuit? (A) 2 amp (B) 4 amp (C) 5 amp (D) 7 amp (E) 9 amp

(A) The 2 A will divide equally between the two branches with 1 A going through each branch. From B to D we have - (1 A)(2 Ω) = -2 V, with B at the higher potential

A 9-volt battery is connected to four resistors to form a simple circuit as shown below. What would be the potential at point B with respect to point D? (A) +2 V (B) +4 V (C) +5 V (D) +7 V (E) +9 V

(C) The 15 Ω resistor would be in parallel with the 30 Ω resistor when the switch is closed.

A battery, an ammeter, three resistors, and a switch are connected to form the simple circuit shown above. When the switch is closed what would happen to the potential difference across the 15 ohm resistor? (A) it would equal the potential difference across the 20 ohm resistor (B) it would be twice the potential difference across the 30 ohm resistor (C) it would equal the potential difference across the 30 ohm resistor (D) it would be half the potential difference across the 30 ohm resistor (E) none of the above

(A) Closing the switch reduces the total resistance of the circuit, increasing the current in the main branch containing bulb 1

A circuit is connected as shown. All light bulbs are identical. When the switch in the circuit is closed illuminating bulb #4, which other bulb(s) also become brighter? (A) Bulb #1 only (B) Bulb #2 only (C) Bulbs #2 and #3 only (D) Bulbs #1, #2, and #3 (E) None of the bulbs.

(B) In series circuits, larger resistors develop more power

A current through the thin filament wire of a light bulb causes the filament to become white hot, while the larger wires connected to the light bulb remain much cooler. This happens because (A) the larger connecting wires have more resistance than the filament. (B) the thin filament has more resistance than the larger connecting wires. (C) the filament wire is not insulated. (D) the current in the filament is greater than that through the connecting wires. (E) the current in the filament is less than that through the connecting wires.

(B) For more light at a given voltage, more current is required, which requires less resistance. R =ρL/A

A junior Thomas Edison wants to make a brighter light bulb. He decides to modify the filament. How should the filament of a light bulb be modified in order to make the light bulb produce more light at a given voltage? (A) Increase the resistivity only. (B) Increase the diameter only. (C) Decrease the diameter only. (D) Decrease the diameter and increase the resistivity. (E) Increase the length only

(E) Before cutting the resistance is R. After cutting we have two wires of resistance ½ R which in parallel is an equivalent resistance of ¼ R. P = V²/R and I = V/R

A length of wire of resistance R is connected across a battery with zero internal resistance. The wire is then cut in half and the two halves are connected in parallel. When the combination is reconnected across the battery, what happens to the resultant power dissipated and the current drawn from the battery? Power Current (A) No change No change (B) Doubles Doubles (C) Quadruples Doubles (D) Doubles Quadruples (E) Quadruples Quadruples

(C) The branch with two 2 Ω resistors has a total resistance of 4 Ω and a potential difference of 12 V. V = IR

An electric circuit consists of a 12 V battery, an ideal 10 A fuse, and three 2 Ω resistors connected as shown above What would be the reading on an ammeter inserted at point B ? (A) 9 A (B) 6 A (C) 3 A (D) 2 A (E) 0 A, since the fuse would break the circuit

(D) With the switch closed, the resistance of the 15 Ω and the 30 Ω in parallel is 10 Ω, making the total circuit resistance 30 Ω and E = IR

An ideal battery, an ideal ammeter, a switch and three resistors are connected as shown. With the switch open as shown in the diagram the ammeter reads 2.0 amperes. When the switch is closed, what would be the current in the circuit? (A) 1.1 A (B) 1.7 A (C) 2.0 A (D) 2.3 A (E) 3.0 A

(A) When the switch is closed, the circuit behaves as if the capacitor were just a wire and all the potential of the battery is across the resistor. As the capacitor charges, the voltage changes over to the capacitor over time, eventually making the current (and the potential difference across the resistor) zero and the potential difference across the capacitor equal to the emf of the battery.

Assume the capacitor C is initially uncharged. The following graphs may represent different quantities related to the circuit as functions of time t after the switch S is closed Which graph best represents the current versus time in the circuit? (A) A (B) B (C) C (D) D (E) E

(B) When the switch is closed, the circuit behaves as if the capacitor were just a wire and all the potential of the battery is across the resistor. As the capacitor charges, the voltage changes over to the capacitor over time, eventually making the current (and the potential difference across the resistor) zero and the potential difference across the capacitor equal to the emf of the battery.

Assume the capacitor C is initially uncharged. The following graphs may represent different quantities related to the circuit as functions of time t after the switch S is closed Which graph best represents the voltage across the capacitor versus time? (A) A (B) B (C) C (D) D (E) E

(A) When the switch is closed, the circuit behaves as if the capacitor were just a wire and all the potential of the battery is across the resistor. As the capacitor charges, the voltage changes over to the capacitor over time, eventually making the current (and the potential difference across the resistor) zero and the potential difference across the capacitor equal to the emf of the battery.

Assume the capacitor C is initially uncharged. The following graphs may represent different quantities related to the circuit as functions of time t after the switch S is closed. Which graph best represents the voltage versus time across the resistor R? (A) A (B) B (C) C (D) D (E) E

(C) Each branch, with two capacitors in series, has an equivalent capacitance of 2 µF ÷ 2 = 1 µF. The three branches in parallel have an equivalent capacitance of 1 µF + 1 µF + 1 µF = 3 µF

Below is a system of six 2-microfarad capacitors. The equivalent capacitance of the system of capacitors is (A) 2/3µF (B) 4/3 µF (C) 3 µF (D) 6 µF (E) 12 µF

(A) A and E failing in the main branch would cause the entire circuit to fail. B and C would affect each other.

Five identical light bulbs, each with a resistance of 10 ohms, are connected in a simple electrical circuit with a switch and a 10 volt battery as shown in the diagram below. Which bulb (or bulbs) could burn out without causing other bulbs in the circuit to also go out? (A) only bulb D (B) only bulb E (C) only bulbs A or E (D) only bulbs C or D (E) bulbs B, C, or D

(C) Shorting bulb 3 decreases the resistance in the right branch, increasing the current through bulb 4 and decreasing the total circuit resistance. This increases the total current in the main branch containing bulb 1.

For the circuit shown, a shorting wire of negligible resistance is added to the circuit between points A and B. When this shorting wire is added, bulb #3 goes out. Which bulbs (all identical) in the circuit brighten? (A) Only Bulb 2 (B) Only Bulb 4 (C) Only Bulbs 1 and 4 (D) Only Bulbs 2 and 4 (E) Bulbs 1, 2 and 4

(E) The total charge to be distributed is +100 µC - 50 µC = + 50 µC. In parallel, the capacitors must have the same voltage so the 20 µF capacitor has four times the charge of the 5 µF capacitor. This gives Q20 = 4Q5 and Q20 + Q5 = 4Q5 + Q5 = 5Q5 = 50 µC, or Q5 = 10 µC

For the configuration of capacitors shown, both switches are closed simultaneously. After equilibrium is established, what is the charge on the top plate of the 5 µF capacitor? (A) 100 µC (B) 50 µC (C) 30 µC (D) 25 µC (E) 10 µC

(D) N is in the main branch, with the most current. The current then divides into the two branches, with K receiving twice the current as L and M. The L/M branch has twice the resistance of the K branch. L and M in series have the same current. Current is related to brightness (P = I²R)

Four identical light bulbs K, L, M, and N are connected in the electrical circuit shown above. In order of decreasing brightness (starting with the brightest), the bulbs are: (A) K = L > M > N (B) K = L = M > N (C) K > L = M > N (D) N > K > L = M (E) N > K = L = M

(D) N is in the main branch, with the most current. The current then divides into the two branches, with K receiving twice the current as L and M. The L/M branch has twice the resistance of the K branch. L and M in series have the same current

Four identical light bulbs K, L, M, and N are connected in the electrical circuit shown above. Rank the current through the bulbs. (A) K > L > M > N (B) L = M > K = N (C) L > M > K > N (D) N > K > L = M (E) N > L = M > K

(D) For no current to flow, the potential drop across R1 must equal the potential drop across R2. For this to occur I1R1 = I2R2. Since the two branches also have the same potential difference as a whole (they are in parallel) we also have I1(R1 + R3) = I2(R2 + R4). Solve for R3

Four resistors, R1, R2, R3, and R4, are connected in the circuit diagram above. When the switch is closed, current flows in the circuit. If no current flows through the ammeter when it is connected as shown, what would be the value of R3?

(C) For the currents in the branches to be equal, each branch must have the same resistance

Given the simple electrical circuit above, if the current in all three resistors is equal, which of the following statements must be true? (A) X, Y, and Z all have equal resistance (B) X and Y have equal resistance (C) X and Y added together have the same resistance as Z (D) X and Y each have more resistance than Z (D) none of the above must be true

(A) I1 is the main branch current and is the largest. It will split into I2 and I3 and since I2 moves through the smaller resistor, it will be larger than I3.

How do the currents I1, I2, and 13 compare? (A) I1 > I2 > I3 (B) I1 > I3 > I2 (C) I2 > I1 > I3 (D) I3 > I1 > I2 (E) I3 > I2 > I1

(B) P = I²R

How much current flows through a 4 ohm resistor that is dissipating 36 watts of power? (A) 2.25 amps (B) 3.0 amps (C) 4.24 amps (D) 9.0 amps (E) 144 amps

(E) For the ammeter to read zero means the junctions at the ends of the ammeter have the same potential. For this to be true, the potential drops across the 1 Ω and the 2 Ω resistor must be equal, which means the current through the 1 Ω resistor must be twice that of the 2 Ω resistor. This means the resistance of the upper branch (1 Ω and 3 Ω) must be ½ that of the lower branch (2 Ω and R) giving 1 Ω + 3 Ω = ½ (2 Ω + R)

If the ammeter in the circuit above reads zero, what is the resistance R ? (A) 1.5 Ω (B) 2Ω (C) 4 Ω (D) 5 Ω (E) 6Ω

(D) If the current in the 6 Ω resistor is 1 A, then by ratios, the currents in the 2 Ω and 3 Ω resistor are 3 A and 2 A respectively (since they have 1/3 and 1/2 the resistance). This makes the total current 6 A and the potential drop across the 4 Ω resistor 24 V. Now use Kirchhoff's loop rule for any branch.

In the accompanying circuit diagram, the current through the 6.0-Ω resistor is 1.0 A. What is the power supply voltage V? (A) 10 V (B) 18 V (C) 24 V (D) 30 V (E) 42 V

(C) Energy = Pt = I²Rt

In the circuit above, the emf's and the resistances have the values shown. The current I in the circuit is 2 amperes. How much energy is dissipated by the 1.5-ohm resistor in 60 seconds? (A) 6 J (B) 180 J (C) 360 J (D) 720 J (E) 1,440 J

(C) S1 must be closed to have any current. Closing S2 will allow current in R2 but closing R3 would short circuit R2

In the circuit above, the resistors all have the same resistance. The battery, wires, and ammeter have negligible resistance. A closed switch also has negligible resistance Closing which of the switches will produce the greatest power dissipation in R 2 ? (A) S 1 only (B) S 2 only (C) S 1 and S 2 only (D) S 1 and S 3 only (E) S 1 , S 2 , and S 3

(E) S1 must be closed to have any current. Closing S3 will short circuit R3, leaving only resistor R1, which is the lowest possible resistance.

In the circuit above, the resistors all have the same resistance. The battery, wires, and ammeter have negligible resistance. A closed switch also has negligible resistance Closing which of the switches will produce the greatest reading on the ammeter? (A) S 1 only (B) S 2 only (C) S 3 only (D) S 1 and S 2 (E) S 1 and S 3

(C) Bulb C in the main branch receiving the total current will be the brightest

In the circuit diagram above, all of the bulbs are identical. Which bulb will be the brightest? (A) A (B) B (C) C (D) D (E) The bulbs all have the same brightness.

(C) The voltage across the capacitor is 6 V (Q = CV) and since the capacitor is in parallel with the 300 Ω resistor, the voltage across the 300 Ω resistor is also 6 V. The 200 Ω resistor is not considered since the capacitor is charged and no current flows through that branch. The 100 Ω resistor in series with the 300 Ω resistor has 1/3 the voltage (2 V) since it is 1/3 the resistance. Kirchhoff's loop rule for the left loop gives E = 8 V.

In the circuit diagrammed above, the 3.00-µF capacitor is fully charged at 18.0 µC. What is the value of the power supply voltage V? (A) 4.40 V (B) 6.00 V (C) 8.00 V (D) 10.4 V (E) 11.0 V

(D) Since there is constant current, bulb 1 remains unchanged and bulbs 2 and three must now split the current. With half the current through bulb 2, the potential difference between A and B is also halved

In the circuit shown above, a constant current device is connected to some identical light bulbs. After the switch S in the circuit is closed, which statement is correct about the circuit? (A) Bulb #2 becomes brighter. (B) Bulb #1 becomes dimmer. (C) All three bulbs become equally brighter. (D) The voltage between points C and D is decreased. (E) The power from the current device is increased.

(D) When the switch has been closed a long time, the voltage across the capacitor is 10 V as the current has stopped and the resistor has no potential drop across it. UC = ½ CV²

In the circuit shown above, the 10 µF capacitor is initially uncharged. After the switch S has been closed for a long time, how much energy is stored in the capacitor? (A) 0 µJ (B) 100 µJ (C) 250 µJ (D) 500 µJ (E) 1000 µJ

(A) When the capacitor is fully charged, the branch with the capacitor is "closed" to current, effectively removing it from the circuit for current analysis

In the circuit shown above, the battery supplies a constant voltage V when the switch S is closed. The value of the capacitance is C, and the value of the resistances are R1 and R2. A long time after the switch has been closed, the current supplied by the battery is (A) V/(R1 + R2) (B) V/R1 (C) V/R2 (D) V(R1 + R2)/R1R2 (E) zero

(D) Utilizing Kirchhoff's loop rile with any loop including the lower branch gives 0 V since the resistance next to each battery drops the 2 V of each battery leaving the lower branch with no current. You can also think of the junction rule where there is 0.04 A going into each junction and 0.04 A leaving to the other battery, with no current for the lower branch.

In the circuit shown above, the current in each battery is 0.04 ampere. What is the potential difference between the points x and y? (A) 8 V (B) 2 V (C) 6 V (D) 0 V (E) 4 V

(D) Kirchhoff's junction rule applied at point X gives 2 A = I + 1 A, so the current in the middle wire is 1 A. Summing the potential differences through the middle wire from X to Y gives - 10 V -(1 A)(2 Ω) = -12 V

In the circuit shown above, the emf's of the batteries are given, as well as the currents in the outside branches and the resistance in the middle branch. What is the magnitude of the potential difference between X and Y? (A) 4 V (B) 8 V (C) 10 V (D) 12 V (E) 16 V

(D) The equivalent resistance of the 20 Ω and the 60 Ω in parallel is 15 Ω, added to the 35 Ω resistor in series gives 15 Ω + 35 Ω = 50 Ω

In the circuit shown above, the equivalent resistance of the three resistors is (A) 10.5 Ω (B) 15Ω (C) 20 Ω (D) 50 Ω (E) 115 Ω

(A) For points a and b to be at the same potential, the potential drop across the 3 Ω resistor must be equal to the potential drop across capacitor C. The potential drop across the 3 Ω resistor is three times the drop across the 1 Ω resistor. For the potential drop across capacitor C to be three times the crop across the 1 µF capacitor, C must be 1/3 the capacitance, or 1/3 µF

In the circuit shown above, the potential difference between points a and b is zero for a value of capacitance C of (A) 1/3 microfarad (B) 2/3 microfarad (C) 2 microfarads (D) 3 microfarads (E) 9 microfarads

(E) The resistance of the two resistors in parallel is r/2. The total circuit resistance is then 10 Ω + ½r, which is equivalent to ε/I = (10 V)/(0.5 A) = 20 Ω = 10 Ω + r/2

In the circuit shown above, the value of r for which the current I is 0.5 ampere is (A) 0 Ω (B) 1 Ω (C) 5 Ω (D) 10 Ω (E) 20 Ω

(B) The total resistance of the 3 Ω and 6 Ω in parallel is 2 Ω making the total circuit resistance 6 Ωand the total current E/R = 1 A. This 1 A will divide in the ratio of 2:1 through the 3 Ω and 6 Ω respectively so the 3 Ω resistor receives 2/3 A making the potential difference IR = (2/3 A)(3Ω) = 2 V.

In the circuit shown above, what is the value of the potential difference between points X and Y if the 6-volt battery has no internal resistance? (A) 1 V (B) 2 V (C) 3 V (D) 4 V (E) 6V

(B) With more current drawn from the battery for the parallel connection, more power is dissipated in this connection. While the resistors in series share the voltage of the battery, the resistors in parallel have the full potential difference of the battery across them

In the diagrams above, resistors R₁ and R₂ are shown in two different connections to the same source of emf εthat has no internal resistance. How does the power dissipated by the resistors in these two cases compare? (A) It is greater for the series connection. (B) It is greater for the parallel connection. (C) It is the same for both connections. (D) It is different for each connection, but one must know the values of R₁ and R₂ to know which is greater. (E) It is different for each connection, but one must know the value of ε to know which is greater.

(A) Current is greatest where resistance is least. The resistances are, in order, 1 Ω, 2 Ω, 4 Ω, 2 Ω and 6 Ω

The batteries in each of the circuits shown above are identical and the wires have negligible resistance. In which circuit is the current furnished by the battery the greatest (A) A (B) B (C) C (D) D (E) E

(E) Current is greatest where resistance is least. The resistances are, in order, 1 Ω, 2 Ω, 4 Ω, 2 Ω and 6 Ω.

The batteries in each of the circuits shown above are identical and the wires have negligible resistance. In which circuit is the equivalent resistance connected to the battery the greatest (A) A (B) B (C) C (D) D (E) E

(B) Since the 5 µF capacitor is in parallel with the battery, the potential difference across it is 100 V. Q = CV

The charge stored in the 5-microfarad capacitor is most nearly (A) 360 µC (B) 500 µC (C) 710 µC (D) 1,100 µC (E) 1,800 µC

(C) When the capacitor is charged, the branch is effectively removed from the circuit, making the circuit a 10 Ω resistor in series with two 10 Ω resistors in parallel. The lone 10 Ω resistor has twice the voltage of the two 10 Ω resistors in parallel with an effective resistance of 5 Ω. The 10 volts will then divide with 3.3 V going to the parallel combination and 6.7 V going to the single 10 Ω resistor. The capacitor is in parallel with the single 10 Ω resistor. Q = CV

The diagram above shows an electrical circuit composed of 3 resistors and 1 capacitor. If each resistor has a resistance of 10 Ω and the capacitor has a value of 10 µF, what would be the charge stored in the capacitor when an EMF of 10 V is maintained in the circuit for a sufficient time to fully charge the capacitor? (A) 23 µC (B) 40 µC (C) 67 µC (D) 100 µC (E) 150 µC

(C) Let the current through the 1 Ω be x. The potential difference across the 1 Ω resistor is then x volts. The current will divide between the upper branch (5 Ω) and the lower branch (9 Ω) with (using the current divider ratio method) 9/(9 + 5) = 9/14 x in the upper branch and 5/14 x in the lower branch. The potential differences are then IR giving for the 2, 3, 4, 5 Ω resistors, respectively 18/14 x, 27/14 x, 20/14 x and 25/14 x volts

The diagram above shows five resistors connected to a voltage source. Which resistor has the greatest potential difference across it? (A) 1 Ω (B) 2 Ω (C) 3 Ω (D) 4 Ω (E) 5 Ω

(D) P = I²R

The diagram below shows five identical resistors connected in a combination series and parallel circuit to a voltage source. Which resistor(s) have the greatest rate of energy dissipation? (A) J only (B) M only (C) N only (D) J&N only (E) K&L only

(B) To retain energy, there must be a capacitor that will not discharge through a resistor. Capacitors in circuits C and E will discharge through the resistors in parallel with them

The five incomplete circuits below are composed of resistors R, all of equal resistance, and capacitors C, all of equal capacitance. A battery that can be used to complete any of the circuits is available. Which circuit will retain stored energy if the battery is connected to it and then disconnected? (A) A (B) B (C) C (D) D (E) E

(E) Even though the wires have different resistances and currents, the potential drop across each is 1.56 V and will vary by the same gradient, dropping all 1.56 V along the same length.

The following diagram represents an electrical circuit containing two uniform resistance wires connected to a single flashlight cell. Both wires have the same length, but the thickness of wire X is twice that of wire Y. Which of the following would best represent the dependence of electric potential on position along the length of the two wires?

(E) The motor uses P = IV = 60 W of power but only delivers P = Fv = mgv = 45 W of power. The efficiency is "what you get" ÷ "what you are paying for" = 45/60

The operating efficiency of a 0.5 A, 120 V electric motor that lifts a 9 kg mass against gravity at an average velocity of 0.5 m/s is most nearly (A) 7% (B) 13% (C) 25% (D) 53% (E) 75 %

(C) When the capacitor is charged, the branch is effectively removed from the circuit, making it a simple parallel circuit. The total resistance is 133.3 Ω and V = IR

Two resistors and a capacitor are connected with a 10 volt battery, a switch and an ideal ammeter to form the simple electrical circuit shown. After the switch is closed and the current in the circuit reaches a constant value, what is the reading on the ammeter in the circuit? (A) 9.2 × 10-2A (B) 8.1 × 10-2A (C) 7.5 × 10-2A (D) 6.9 × 10-2A (E) zero

(E) Since these resistors are in series, they must have the same current.

Two resistors of the same length, both made of the same material, are connected in a series to a battery as shown above. Resistor II has a greater cross. sectional area than resistor I. Which of the following quantities has the same value for each resistor? (A) Potential difference between the two ends (B) Electric field strength within the resistor (C) Resistance (D) Current per unit area (E) Current

(A) P = I²R and the current is the same through each resistor.

Two resistors, one with resistance R and the second with resistance 4R are placed in a circuit with a voltage V. If resistance R dissipates power P, what would be the power dissipated by the 4R resistance? (A) 4 P (B) 2 P (C) P (D) 1/2 P (E) 1/4 P

(D) Resistance of the 2000 Ω and 6000 Ω in parallel = 1500 Ω, adding the 2500 Ω in series gives a total circuit resistance of 4000 Ω. I_total = I_1 = E/R total

What is the current I1? (A) 0.8 mA (B) 1.0 mA (C) 2.0 mA (D) 3.0 mA (E) 6.0 mA

(A) If you perform Kirchhoff's loop rule for the highlighted loop, you get a current of 0 A through the 6 Ω resistor.

What is the current through the 6.0 Ω resistor shown in the accompanying circuit diagram? Assume all batteries have negligible resistance. (A) 0 (B) 0.40 A (C) 0.50 A (D) 1.3 A (E) 1.5 A

(D) P = V²/R

What is the resistance of a 60 watt light bulb designed to operate at 120 volts? (A) 0.5 Ω (B) 2 Ω (C) 60 Ω (D) 240 Ω (E) 7200 Ω

(C) For the 3 capacitors in series on the right C_T = C/3. Adding to the capacitor in parallel gives C + C/3 = 4C/3

What would be the equivalent capacitance of the circuit shown if each capacitor has a capacitance of C? (A) ¼ C (B) ¾ C (C) 4/3 C (D) 3C (E) 4C

(D) On the right, the 6 Ω and 3 Ω resistor in parallel have an equivalent resistance of 2 Ω. Added to the 4 Ω resistance in the middle branch which is in series with the pair gives 6 Ω across the middle. This is in parallel with the 3 Ω resistor at the top giving an equivalent resistance of 2 Ω. Lastly add the 4 Ω resistor in the main branch giving a total circuit resistance of 6 Ω. V = IR.

What would be the total current being supplied by the battery in the circuit shown above? (A) 3.0 amperes (B) 2.25 amperes (C) 2.0 amperes (D) 1.5 amperes (E) 1.0 amperes

(E) by definition of a parallel circuit

When any four resistors are connected in parallel, the _______ each resistor is the same. (A) charge on (B) current through (C) power from (D) resistance of (E) voltage across

(E) P = V²/R

When lighted, a 100-watt light bulb operating on a 110-volt household circuit has a resistance closest to (A) 10-2 Ω (B) 10-1 Ω (C) 1 Ω (D) 10 Ω (E) 100 Ω