Circuits

(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

(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

(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.

(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) 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

(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) 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

(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 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

(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) 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

(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?

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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) 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

(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

(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

(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

(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 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

(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) 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

(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

(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

(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

(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

(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.

(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.

(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

(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

(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

(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

(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

(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

(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

(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.

(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.

(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

(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

(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

(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?

(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

(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

(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

(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

(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

(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

(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

(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) 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

(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

(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.

(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

(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

(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) 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.