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AP Physics C Electricity : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C Electricity

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1 Diagnostic Test 46 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Capacitors

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

Possible Answers:

$C_{eq}= 2C$

$C_{eq}=\frac{C}{2}$

$\frac{C}{2} < C_{eq} < C$

$C < C_{eq}< 2C$

$C_{eq}=C$

Correct answer:

$C_{eq}=\frac{C}{2}$

Explanation:

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

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Example Question #4 : Capacitors

Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?

Possible Answers:

$\frac{eQ \epsilon _o A}{d^2}$

$\frac{eQ}{\epsilon _o A d}$

$\frac{eQd}{\epsilon _oA}$

$\frac{e \epsilon _o A}{Qd}$

$\frac{eQ\epsilon _o A}{d}$

Correct answer:

$\frac{eQd}{\epsilon _oA}$

Explanation:

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

$C = \frac{Q}{V}$

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

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Example Question #3 : Electric Circuits

A $5 \mu F$ and $10 \mu F$ capacitor are connected in series with a 12V battery. What is the magnitude of the charge on one plate of the $5 \mu F$ capacitor?

Possible Answers:

$40 \mu C$

$180 \mu C$

$2.4 \mu C$

$10 \mu C$

$60 \mu C$

Correct answer:

$40 \mu C$

Explanation:

Relevant equations:

Q = CV

$\frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2}$

First, find the equivalent capacitance of the whole circuit:

$\frac{1}{C_{eq,series}}= \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$

$C_{eq, series} = \frac{10}{3} \mu F$

Use this equivalent capacitance to find the total charge:

$Q = CV = (\frac{10}{3} \mu F) * (12 V) = 40 \mu C$

Capacitors in series always have the same charge on each unit, so the charge on the $5 \mu F$ capacitor is also $40 \mu C$ .

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Example Question #1 : Understanding Resistors

Resistors are one of the most important basic components of a circuit. With very few exceptions, all circuits have at least one kind of resistor component. An ammeter is a device that measures current flowing through a circuit. Ammeters are always connected to a circuit in series.

Which of the following accurately explains why ammeters must be connected in series within a circuit, and never in parallel?

Possible Answers:

An ammeter connected in parallel would give high readings of current because the current would be flowing through both the regular current path and the ammeter, but disproportionately through the ammeter

An ammeter connected in parallel would give very low readings of current because current would be flowing through both the regular current path and the ammeter, but disproportionately through the regular path

An ammeter connected in parallel would not read any current because the current would have no driving force through the ammeter

An ammeter connected in parallel would give the current flowing through the circuit a pathway with minimal resistance, creating a very large current that could harm the ammeter

An ammeter connected in parallel would give readings of half the current flowing through the circuit because current would flow evenly between the ammeter and the circuit's regular pathway

Correct answer:

An ammeter connected in parallel would give the current flowing through the circuit a pathway with minimal resistance, creating a very large current that could harm the ammeter

Explanation:

In order to give accurate readings of current, ammeters have very low resistances. If connected in parallel, the voltage pushing current through the circuit would push a very strong current through the ammeter and virtually no current through the circuit's regular path. This would not only lead to a bad reading of current, but more often than not a broken ammeter. This phenomenon is an indication of why resistors are so important: they limit the current such that a circuit does not exceed its current-carrying capacity.

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Example Question #1 : Understanding Resistors

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

Why was the potential of the battery measured differently when the wire was connected?

Possible Answers:

The wire has a slight internal resistance and caused a potential drop

Once the current is distributed across the wire, there is less charge per unit length of the circuit and so the voltage is lower

The wire connected to the battery must be connected to something else where this potential is being lost

The wire is causing electrons to be lost to the air, which lowers the potential measured

The difference in potential measured is small enough to be disregarded

Correct answer:

The wire has a slight internal resistance and caused a potential drop

Explanation:

All wires have at least some internal resistance. The most likely explanation for this is that the wire is displaying slight resistance, and therefore caused the measured potential to be less than it was before.

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Example Question #1 : Resistors

Three resistors and a battery form the following circuit.

Circuit_1

Calculate the equivalent resistance if the resistors have the following values.

$R_1=40\Omega$

$R_2=25\Omega$

$R_3=38\Omega$

Possible Answers:

$R_{123}=24.5\Omega$

$R_{123}=92.3\Omega$

$R_{123}=45.2\Omega$

$R_{123}=2.5\Omega$

$R_{123}=134.5\Omega$

Correct answer:

$R_{123}=24.5\Omega$

Explanation:

First, calculate the equivalent resistance of R_2 and R_3 . Since these two resistors are arranged in series, we just take the sum of their values.

$R_{23}=R_{2}+R_{3}$

$R_{23}=25\Omega+38\Omega$

$R_{23}=63\Omega$

With resistors 2 and 3 combined together in a single value, the following circuit is formed.

Circuit_2

Notice that and $R_{23}$ are arranged in parallel. To calculate the equivalent resistance of this parallel pair, we use the following equation.

$\frac{1}{R_{123}}=\frac{1}{R_1}+\frac{1}{R_{23}}$

Plug in the values, and solve for $R_{123}$ .

$\frac{1}{R_{123}}=\frac{1}{40\Omega}+\frac{1}{63\Omega}$

$R_{123}=24.5\Omega$

This is just like the circuit shown below.

Circuit_3

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Example Question #1 : Current

A lamp has a 60W bulb. If the house wiring provides 110V to light up that bulb, how much current does the bulb draw?

Possible Answers:

$0.72\ \text{A}$

$1.37\ \text{A}$

$0.03\ \text{A}$

$0.55\ \text{A}$

$5.83\ \text{A}$

Correct answer:

$0.55\ \text{A}$

Explanation:

The formula for power is , P=IV and we are given the following values.

$P=60\ \text{W}$

$V=110\ \text{V}$

Solve for the current, .

$I=\frac{P}{V}=\frac{60\ \text{W}}{110\ \text{V}}=0.55\ \text{A}$

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Example Question #2 : Current

A particle accelerator with a radius of 500 meters can have up to $10^{15}$ protons circulating within it at once.

$q_{proton}=1.60*10^{-19}C$

How fast must the protons in the accelerator move in order to produce a current of 1A?

Possible Answers:

$9.21*10^{7}\frac{m}{s}$

$2.04*10^8\frac{m}{s}$

$1.96*10^7\frac{m}{s}$

$1.95*10^6\frac{m}{s}$

$3.00*10^8\frac{m}{s}$

Correct answer:

$1.96*10^7\frac{m}{s}$

Explanation:

The current produced is the total charge that circulates the particle accelerator per unit time.

$I=\frac{dQ}{dt}$

We calculate this by the equation:

$I = \frac{nqv}{2\pi R}$

is the number of protons, is the charge per proton, is the velocity of each proton, and is the radius of the particle accelerator.

Using the given current, we then solve for the velocity:

$1A = \frac{(10^{15})(1.60*10^{-19}C)v}{2\pi (500m)}$

$v=\frac{(1A)2\pi(500m)}{(10^{15})(1.60*10^{-19}C)}= 1.96*10^7\frac{m}{s}$

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Example Question #221 : Ap Physics C

Three resistors R₁, R₂, and R₃ and the capacitor, C, are connected to an ideal battery V to complete the circuit as shown.

Ps0_capacitorcircuit

After the circuit has been connected for a very long time, the currents in each branch of the circuit are measured to be some values I_1 , , and , and the charge on the capacitor some value .

Which of the following equations is a correct expression regarding the voltage of the circuit after a long time?

Possible Answers:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{2}I_{2}$

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

$0=-V-R_{2}I_{2} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} -R_{2}I_{2}$

Correct answer:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

Explanation:

Identifying the sum of the voltage drops and rises (Kirchoff's Loop Law) around the three possible loops of this circuit is the key to answering this question correctly. The following signs can be assigned to each of the circuit elements based on the direction of the currents given.

Ps0_capacitorcircuit_2

Use Ohm's law and the equation for capacitance to derive terms for the voltage across each element of the circuit.

V=IR

$C=\frac{Q}{V}\rightarrow V=\frac{Q}{C}$

There are three possible paths through the circuit, resulting in three correct equations that could be derived:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} +R_{2}I_{2}$

$0=+V-R_{2}I_{2} -R_{1}I_{1}$

Only one of the given answer options matches up correctly to these.

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Example Question #1 : Circuit Relationships

A simple circuit consists of a $2\ \text{M}\Omega$ resistor that's connected to a battery. How much power is dissipated by the resistor?

Possible Answers:

$4*10^{-5}W$

$5.5*10^{-6}W$

$8*10^{-4}W$

$2*10^{-2}W$

$7.2*10^{-2}W$

Correct answer:

$4*10^{-5}W$

Explanation:

One of the formulas for power is $P=\frac{V^2}{R}$ . We are given the values of the voltage and resistance.

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=2*10^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{2*10^6\ \Omega}$

$P=4*10^{-5}\ \text{W}$

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AP Physics C Electricity : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #4 : Capacitors

Example Question #4 : Capacitors

Example Question #3 : Electric Circuits

Example Question #1 : Understanding Resistors

Example Question #1 : Understanding Resistors

Example Question #1 : Resistors

Example Question #1 : Current

Example Question #2 : Current

Example Question #221 : Ap Physics C

Example Question #1 : Circuit Relationships

All AP Physics C Electricity Resources

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