Physics Final Exam: Test 2 Material Review (CIRCUITS)

(A: find the value of R using R = rho*length/area, which equals 2.987 ohms, then using the given voltage of 9, current can be found using Ohm's law; I = 3.013 A) (B: find the new value of R due to the temperature change with the following equation R = Rnot*[1+alpha(T- Tnot)], which gives R=3.1 ohms, which still using the same voltage, I can be calculated with Ohm's law which gives I = 2.903 A)

(A) A 34.5-m length of copper wire at 20.0 degrees C has a radius of 0.25 mm. If a potential difference of 9.00 V is applied across the length of the wire, determine the current in the wire. (B) If the wire is heated to 30.0 degrees Celsius the 9.00-V potential difference is maintained, what is the resulting current in the wire? Note the following: alpha = 3.9x10^-3 C^-1; resistivity of copper = 1.7x10^-8.

(Note: ∆V = ∆Vmax*[1-e^-t/RC], plug in the knonw values and solve for R) (Answer: 587 kiloOhms)

A 10.0 microF capacitor is charged by a 10-V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V in a time interval of 3.00 s after charging begins. Find R.

D; D (part A: V = IR = 3*6 = 18V) (part B: E = V + IR or I(R+r), which when solved for r gives 2 ohms, note this is an unrealistically large internal resistance)

A battery of emf 24 V is connected to a 6-ohm resistor. As a result, current of 3A exists in the resistor. The terminal potential difference of the battery is: A) 0 B) 6 V C) 12 V D) 18 V E) 24 V What is the "internal resistance" of this battery (in ohms)? A) 0.0 ohms B) 0.5 ohms C) 1.0 ohms D) 2.0 ohms

C (note the equation q charging = Qmax*[1-e^-t/RC], and we know that q = 0.5 of the max, so 0.5 = 1-e^-t/RC, and we use this to solve for C, which comes to 20 microF)

A certain capacitor, in series with a 720-ohm resistor, is being charged. At the end of 10 ms its charge is half the final value. The capacitance is about: A) 9.6 microF B) 14 microF C) 20 micro F D) 7.2 microF E) 10 microF

(based on the chart we are trying to solve for resistivity: given voltage and current, which can lead us to resistance by Ohm's law, and then by this equation R = rho*l/area we can find the resistivity, and then using the table find the specific metal we are looking for) (answer; silver)

A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9,11 V, and the current is found to be 36.0 A. Assume a temperature of 20.0 degrees C and, using the following table, identify the metal out of which the wire is made.

(A: note C1 and C2 are in series in both places, and both of these equivalents are in parallel with C3; C2 and C2 in the bottom are in parallel, and this equivalent is in series with the top part of the circuit; overall this gives a Ceq = 6.05 microF) (B: find the Qtotal using Ceq and 60V, which gives 3.63x10^-4 C. Using this value find the ∆V value of just the top part of the circuit, which is Q total/Ceq which equals 41.87 V. This is the voltage drop across C3, and using the capacitance we can find the charge at this capacitor, which is equal to 83.7 microC)

A) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in the figure. Take C1 = 5 microF, C2 = 10 microF, and C3 = 2 microF. B) What charge is stored on C3 if the potential difference between points a and b is 60.0 V?

(A: C = A*epsilon not/d = 2.3cm^2*1 m^2/100^2 cm^2 *8.85x10^-12 divided by 1.5x10^-3 = 1.357 picoF) (B: Q = C*V = 16.3 picoC) (C: E = V/d = 12/1.5x10^-3 = 8000 N/C or V/m)

An air-filled parallel-plate capacitor has plates of area 2.3cm^2 separated by 1.5 mm. A) Find the value of its capacitance. B) The capacitor is connected to a 12.0 V battery. What is the charge on the capacitor? C) What is the magnitude of the uniform electric field between the plates?

(Note: power = current*voltage = energy/time, where energy is heat = mc∆T; can set time = energy divided by current*voltage) (the boiling point of water is the final temperature, or 100 degrees Celsius, so the energy = 4186*[100-21]*2.4 = 793665.6 J) (Solving for time: t = 793665.6/120*8.5 = 778.104 seconds = about 13 minutes)

An automatic coffee maker uses a resistive heating element to boil 2.4 kg of water that was initially at 21 degrees C. The current delivered to the coffee pot is 8.5 A when it is plugged into a 120 V electrical outlet. If the specific heat capacity of water is 4186 J/kg*C approximately how long does it take to bring the water to its boiling point?

(Part A: tau = RC = 0.002 s) (Part B: Qmax = capacitance*epsilon = 1.8x10^-4 coulombs) (Part C: q of t = Qmax*[1-e^-t/RC] = 1.14 x10^-4 coulombs, when plugged in)

An uncharged capacitor and a resistor are connected in series to a source of emf. If epsilon = 9.00 V, C = 20.0 microF, and R = 100 ohms, find (a) the time constant of the circuit, (b) the maximum charge on the cpacitor, and (c) the charge on the capacitor at a time equal to one time constant after the battery is connected.

1.1*10^-2 (U = 0.5C*V^2 = 0.5*1x10^-6*150^2)

Consider a 1.0 microF parallel plate capacitor with a plate spacing of 50 microm is charged so that the field in the air between the plates is 3.0 x 10^6 V/m. How much energy is stored in this capacitor?

discharge

Consider the following equation: C = k*(epsilon not *A)/d. In theory, d could be made very small to create a very large capacitance. IN practice, there is a limit to d. d is limited by the electric _____ that could occur through the dielectric medium separating the plates.

B (voltage across each resistor is the same since they are connected in parallel, so divide 20V by 20 ohms for each one, which gives a current of 1.0 A)

Four 20 ohm resistors are connected in parallel and the combination is connected to a 20-V emf device. The current in any one of the resistors is: A) 0.25 A B) 1.0 A C) 4.0 A D) 5.0 A E) 100 A

A (they are in series: so need to find equivalent resistance, which is 80 ohms and divide this number from 20-V, which gives 0.25 amps)

Four 20 ohm resistors are connected in series and the combination is connected to a 20-V emf device. The current in any one of the resistors is: A) 0.25 A B) 1.0 A C) 4.0 A D) 5.0 A E) 100 A

D (current in = current out, here current in equals 4, but current out equals 7, so we need to add 3amps to the current going in so the junction rule is satisfied)

Four wires meet at a junction. The first carries 4 A into the junction, the second carries 5 A out of the junction, and the third carries 2A out of the junction. The fourth carries: A) 7A out of the junction B) 7A into the junction C) 3A out of the junction D) 3A into the junction E) 1A into the junction

voltage across; charge

If the capacitor remains connected to a battery, the ________ the capacitor necessarily remains the same. If the capacitor is disconnected from the battery, the capacitor is an isolated system and the ______ remains the same.

A (note the equation U = 0.5*C*∆V^2)

If the potential difference ∆V across a capacitor is doubled, what happens to the energy stored? A) it becomes four times larger B) it becomes two times larger C) it is unchanged D) it becomes one-half as large E) it becomes one-fourth as large

B (A is false because in series the equivalent capacitance is smaller than each individual one because 1/Ceq = 1/C1 + 1/C2; C is false because ∆V = Q/C, so they are inversely proportional; lastly D and E are false because the charges on each capacitor in series will be the same)

If we have three unequal capacitors, initially uncharged, which are connected in series across a battery, which of the following statements is true? A) the equivalent capacitance is greater than any of the individual capacitances B) the largest voltage appears across the smallest capacitance C) the largest voltage appears across the largest capacitance D) the capacitor with the largest capacitance has the greatest charge E) the capacitor with the smallest capacitance has the smallest charge

epsilon; internal resistance (within the circuit)

In real (non ideal) batteries ______ represents the maximum voltage that can be delivered, and r represents the _________.

A (when the key is pressed, the plate separation is decreased, and the capacitance increases; capacitance depends only on how the capacitor is constructed and not on the external circuit)

Many computer keyboard buttons are constructed of capacitors as shown in the figure. When a key is pushed down, the soft insulator between the movable plate and the fixed plate is compressed. When the key is pressed, what happens to the capacitance? A) it increases B) it decreases C) it change in a way you cannot determined because the electric circuit connected to the keyboard button may cause a change in ∆V

(A: connecting plates of like sign places the capacitors in parallel, so the voltage on each capacitor remains the same: U total = 0.5*C*∆V^2 + 0.5*C*∆V^2 = C∆V^2) (B: C = epsilon not *A/d, the altered capacitor has new capacitance = C/2, and this change in capacitance results in a new potential difference ∆V' across the parallel capacitors; we can solve for the new potential difference because the total charge remains the same: 2Q = C∆V + C∆V = C∆V' + C/2*∆V'--> ∆V' = 4∆V/3) (C: each capacitor has potential difference ∆V'; U' total = 0.5*C*∆V'^2 + 0.5*C'*∆V'^2 = 0.5*C*[4/3*∆V]^2 + 0.5*C/2*[4/3*∆V]^2 = 12/9*C*∆V^2 = 4C*∆V^2/3) (D: positive work is done by the agent pulling the plates apart)

Two identical parallel-plate capacitors, each with capacitance C, are charged to potential difference ∆V and then disconnected form the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. A) Find the total energy of the system of two capacitors before the plate separated is doubled. B) Find the potential difference across each capacitor after the plate separation is doubled C) Find the total energy of the system after the plate separation is doubled D) Reconcile the difference in the answers in parts a and c with the law of conservation of energy

C (current prefers the path of least resistance, but it will still pass through both, just less will go through R1)

What happens when the switch is closed? A) the current chooses the pathway of least resistance and only goes through R2 B) the current splits with half now going through each resistor C) current passes through both resistors but more goes through R2 D) we need to know the voltage of the battery to answer this question

E (drift speed depends on current, which is dependent on movement of positive charge, which opposes the movement of electrons, when the current decreases the drift speed decreases)

What is the drift speed of an electron in a typical wire? A) 2.99x10^8 m/s (speed of light) B) 3.40 x 10^2 m/s (speed of sound) C) 1 m/s (human walking pace) D) 1.3x10^-2 m/s (pace of a snail) E) 2.5x10^-4 m/s (pace of a very slow slug)

C (they are connected in series, so the current is the same at all points in the circuit)

When the switch is closed, how does the current compare at points A and B? A) current at point A> current at point B B) current at point A < current at point B C) Current at point A = current at point B D) need more information to answer

increases

With a dielectric, the capacitance becomes C = k*Cnot. The capacitance ________ by the factor k when the dielectric completely fills the region between the plates. K is the dielectric contstant of the material.

B (for a given voltage, the energy stored in a capacitor is proportional to C, so in order to maximize equivalent capacitance, the capacitors should be connected in parallel)

You have three capacitors and a battery. IN which of the following combinations of the three capacitors is the maximum possible energy stored when the combination is attached to the battery? A) series B) parallel C) no difference because both combinations store the same amount of energy

dielectric

a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance (ex. include rubber, glass, and waxed paper)

E (tau = RC = 720*20x10^-6 = 0.0144 s = 14.4 ms)

A certain capacitor, in series with a resistor, is being charged. The capacitor has a value of 20x10^-6, and the resistor has a value of 720 ohms. At the end of 10 ms its charge is half the final value. The time constant for the process is about: A) 0.43 ms B) 2.3 ms C) 6.9 ms D) 10 ms E) 14 ms

potential difference (∆V = E*I, which eventually yields Ohm's Law of V=I*R)

A potential difference ∆V = Vb-Va maintained across the conductor sets up an electric field E, and this field produces a current I that is proportional to the ______.

dielectric strength

For a given d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the ______ of the material.

R = Rnot[1+alpha(T-Tnot)] (where R not is the resistance at T not and alpha, the temperature coefficient of resistivity in units of Celsius to the -1)

Resistance does depend on temperature. This is the equation? ______.

(Equations: from junction rule i1 + i2 = i3 = 2, and from the loop rule 15 = 7i1 + 5i3 = 7*i1 + 5*[i1 + i2] = 12*i1 + 5*i2) (Solving for i1: solve system of equations by multiplying -5 by i1 + i2 = 2, giving -5*i1 - 5*i2 = -10, and add this equation to 15 = 12*i1 + 5*i2, which gives i1 = 0.714A) (Solving for i2: plug i1 = 0.714 into i1 + i2 = 2, which gives i2 = 1.286 amps) (solving for epsilon: epsilon = 2*5 + 2*1.286 = 12.57 V)

The ammeter shown in the figure reads 2A. Find A) I1 B) I2 and C) epsilon

positive

The direction of the current is the direction in which ______ charges flow when free to do so.

(A: area = 2.00 cm^2 = 2x10^-4 m^2; take the derivative of the charge equation, which gives 12t^2+5, and plug in 1 second for time; this gives I = 17.0A) (B: charge density J = I/A = 17/2x10^-4 = 85 kA/m^2)

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00cm^2 varies with time according to the equation q=4t^3 + 5t + 6, where t is in seconds. (A) What is the instantaneous current through the surface at 5=1.00 s? (B) What is the value of the current density?

A (in series, the inverses of the capacitances add, so the equivalent capacitance is smaller than each individual capacitance)

Two capacitors are identical. They can be connected in series or in parallel. If you want the smallest equivalent capacitance for the combination, how should you connect them? A) in series B) in parallel C) either way because both combinations have the same capacitance

(B: U = 0.5*C*∆V^2; where C = C1 + C2 = 30 microF; U = 0.5*30x10^-6*100^2 = 0.150 J) (C: 1/Ctotal = 1/C1 + 1/C2; Ceq = 4.17 microF) (C: U = 0.5*C*∆V^2 --> ∆V = 268V)

Two capacitors, C1 = 25 microF and C2 = 5.0 microF, are connected in parallel and charged with a 100-V power supply. A) Draw a circuit diagram B) calculated the total energy stored in the two capacitors. C) What if? What potential difference would be required across the same two capacitors connected in series for the combination to store the same amount of energy as in part b? D) Draw a circuit diagram o the circuit described in part c.

(because the circuit is in series, the current is the same at all points, so only the resistances are changing: light A, because it is brighter has a lower resistance, and therefore a lower potential difference than at B)

What can you say about the potential difference across light A? Which light (A or B) has the greater resistance?

Ceq = C1+C2 (where Q total = Ceq*∆V) 1/Ceq = 1/C1 + 1/C2 (where Q1=Q2, and ∆V total = Q1/C1 + Q2/C2)

What does Ceq equal for parallel capacitors? What does Ceq equal for capacitors connected in series?

C (because it has not split yet, so this current will stay the same)

What happens to the current passing through the ammeter "A" when the switch is closed? A) the amount of current increases B) the amount of current decreases C) the amount of current stays the same D) need more information