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AP Physics 2 : AP Physics 2

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Example Questions

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Example Question #39 : Electricity And Magnetism

What is the resistance in a wire carrying a voltage of 10V and with a power of ?

Possible Answers:

$6.25\Omega$

$16\Omega$

$25\Omega$

$1.6\Omega$

$2.5\Omega$

Correct answer:

$25\Omega$

Explanation:

The formula we can use here is the power formula that involved both resistance and voltage:

$P=\frac{V^2}{R}$

We are given the voltage and power, allowing us to solve for the resistance.

$4W=\frac{(10V)^2}{R}$

$R=\frac{100V^2}{4W}$

$R=25\Omega$

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Example Question #1 : Ohm's Law

A $\small 12V$ battery supplies power to a circuit with a current of $\small 16A$ . What is the resistance in this circuit?

Possible Answers:

$0.75\Omega$

$192\Omega$

$9\Omega$

$12\Omega$

$1.33\Omega$

Correct answer:

$0.75\Omega$

Explanation:

To find resistance from voltage and current, we will need to use Ohm's law:

V=IR

$R= \frac{V}{I}$

We are given the voltage and current, allowing us to solve for the resistance.

$R = \frac{12V}{16A} = 0.75\Omega$

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Example Question #60 : Circuits

An electrician wishes to cut a copper wire $(\rho = 1.724 * 10^{-8}\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Possible Answers:

38m

960m

10cm

2.6cm

9.6m

Correct answer:

960m

Explanation:

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = \frac{\rho L}{A}$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = \frac{RA}{\rho }$ . We must first solve for A using the radius, 0.725mm.

$A = \pi r^{2} = \pi (0.000725m)^{2}=1.65 * 10^{-6}m^{2}$

Plugging in our numbers gives the answer, 960m.

$L = \frac{10\Omega * 1.65 * 10^{-6}m^{2}}{1.724*10^{-8}\Omega m} = 957m \approx960m$

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Example Question #1 : Resistors And Resistance

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

What is the net resistance of the circuit?

Possible Answers:

$11\Omega$

$2.5\Omega$

$4\Omega$

$9\Omega$

$2\Omega$

Correct answer:

$4\Omega$

Explanation:

First we need to combine the resistors in parallel.

$\frac{1}{R_{A4}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{A4}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

Now that we have simplified the two resistors in parallel, we need to determine how the R_eq of the two parallel resistors and the resistance due to R₁ are arranged. We can see that these resistors are now in series, thus we can directly add their resistances together to get the overall resistance of the circuit.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Remember as a general rule, the equivalent resistance of resistors in series is an arithmetic sum of their individual resistances.

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Example Question #1 : Circuits

Screen_shot_2013-10-09_at_11.08.15_pm

Instead of assuming that the 9V battery has no internal resistance, what would the terminal potential of the battery be if it has an internal resistance of $0.9\Omega$ ?

Possible Answers:

8.1V

2.394V

6.975V

4.643V

Correct answer:

6.975V

Explanation:

Remember that internal resistance lowers the EMF of the battery. We need to calculate the voltage drop that would occur inside the battery, and then subtract that from the “stated” voltage to determine the terminal potential of the battery.

We first need the current supplied by the battery. We can use the formula V = IR because we have the voltage drop across the circuit (9V) and can calculate the equivalent resistance.

$\frac{1}{R_{A4}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{A4}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Now, using V=IR, we can find the current in the circuit.

V = IR

I = V/R = 9V/4Ω = 2.25A

Plugging in this value with the internal resistance of the battery gives us the voltage decrease.

V = IR = (2.25A)(0.9Ω) = 2.025V

Terminal Potential = 9V – 2.025V = 6.975V

As we can see, battery companies are interested in keeping the internal resistance of batteries at a minimum, because it lowers the overall terminal potential of the battery.

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Example Question #1 : Capacitors And Dielectrics

Capacitors with capacitances of 3 μF, 7 μF and 10 μF are wired in parallel. What is the capacitance of the circuit?

it cannot be determined without knowing the resistance of the circuit
it cannot be determined without knowing the time constant in the circuit
20 μF
approximately 1.75 μF
none of these is correct

Possible Answers:

Correct answer:

Explanation:

Response 3 is the correct choice. Electrons will space themselves as far apart as possible, because of charge repulsion; therefore, in a parallel arrangement, they will jump onto each capacitor and “load it up.” The parallel capacitance is calculated by simply adding the individual values. The value would be about 1.75 μF if the three were connected in series, where the formula is $\dpi{100} \small \frac{1}{C_{total}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$ (reciprocal of total equals sum of reciprocal of each). The time constant of a circuit is given by the formula τ = RC, and it relates to how fast a capacitor can charge to full capacitance.

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Example Question #2 : Capacitors And Dielectrics

A $2\mu F$ and $6\mu F$ capacitor are connected in series with a 12V battery. What is the maximum charge stored on the $2\mu F$ capacitor?

Possible Answers:

$18\mu C$

$24 \mu C$

$96 \mu C$

$13.5 \mu C$

$4.5 \mu C$

Correct answer:

$18\mu C$

Explanation:

First find the equivalent capacitance of the two capacitors by adding their inverses.

$\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{6}$ $\rightarrow C_{eq} = \frac{6}{4} = 1.5\mu F$

Then, we can find the charge stored on this equivalent capacitor.

$Q_{eq}=C_{eq}V = 1.5 * 12 = 18\mu C$

For capacitors in series the charges on each must be equal, and also equal to the charge on the equivalent capacitor. The answer is $18\mu C$ .

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Example Question #1 : Capacitors And Dielectrics

Batteries and AC current are often used to charge a capacitor. A common example of capacitor use is in computer hard drives, where capacitors are charged in a specific pattern to code information. A simplified circuit with capacitors can be seen below. The capacitance of C₁ is 0.5 μF and the capacitances of C₂ and C₃ are 1 μF each. A 10 V battery with an internal resistance of 1 Ω supplies the circuit.

Pretext Question_2

How is the charge stored on the capacitor?

Possible Answers:

Unevenly distributed on the capacitor surface

Unevenly distributed inside the capacitor

Evenly distributed on the capacitor surface

Evenly distributed inside the capacitor

Correct answer:

Evenly distributed on the capacitor surface

Explanation:

Charge is evenly distributed on the surface of the capacitor. If we think back to electric force, we know that positive charges repel other positive charges and negative charges repel other negative charges; thus, the charges are evenly distributed to minimize the force between them. We can see how this looks in diagrammatic form below.

Charge_dist

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Example Question #41 : Circuits

Pretext Question_2

What is the equivalent capacitance of C₂ and C₃?

Possible Answers:

0.5μF

0.33μF

3μF

2μF

Correct answer:

0.5μF

Explanation:

First, we need to determine how these capacitors are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals:

$\frac{1}{C_{eq}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C_eq = 0.5μF

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Example Question #5 : Capacitors And Dielectrics

Pretext Question_2

What is the equivalent capacitance of the circuit?

Possible Answers:

3μF

2μF

4μF

1μF

Correct answer:

1μF

Explanation:

First, we need to determine how capacitors C₂ and C₃ are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals.

$\frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C₂₃ = 0.5μF

Next, we need to determine how we can find the C_eq by simplifying C₂₃ and C₁. We can see that C_eq and C₁are in parallel, thus we can directly add the individual capacitances.

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #39 : Electricity And Magnetism

Example Question #1 : Ohm's Law

Example Question #60 : Circuits

Example Question #1 : Resistors And Resistance

Example Question #1 : Circuits

Example Question #1 : Capacitors And Dielectrics

Example Question #2 : Capacitors And Dielectrics

Example Question #1 : Capacitors And Dielectrics

Example Question #41 : Circuits

Example Question #5 : Capacitors And Dielectrics

All AP Physics 2 Resources

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