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AP Physics 2 : Circuits

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Example Questions

Example Question #9 : Capacitors And Electric Fields

A parallel plate capacitor with a separation of $.5\textup{ mm}$ and surface area $150\textup{ cm}^2$ is in series with a $9\textup{ V}$ battery. How will the electric field in the capacitor change if the separation is doubled?

Possible Answers:

Tripled

Doubled

Quadrupled

It will stay the same

Halved

Correct answer:

Halved

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

V_b=V_c

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

V_c=E*d

Combing equations and solving for

$E=\frac{V_b}{d}$

From this, it can be seen that doubling the separation will halve the electric field.

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Example Question #10 : Capacitors And Electric Fields

Possible Answers:

No change

Impossible to determine

Doubled

Halved

Tripled

Correct answer:

Doubled

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

V_b=V_c

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

V_c=E*d

Combing equations and solving for

$E=\frac{V_b}{d}$

From this, it can be seen that doubling the voltage of the battery will doubled the electric field inside the capacitor.

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Example Question #31 : Circuit Components

A capacitor is placed in series with a battery. The plates are 1mm apart and each have a 4cm^2 surface area. Determine the charge on the positive plate.

Possible Answers:

None of these

$3.95*10^{-11}C$

$4.54*10^{-11}C$

$8.44*10^{-11}C$

$2.13*10^{-11}C$

Correct answer:

$2.13*10^{-11}C$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$

Combining equations:

V_1=E*d

Solving for :

$\frac{V_1}{d}=E$

Converting to and plugging in values:

$\frac{V_1}{d}=E$

$6000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$

Combining equations:

$E*\epsilon_0*A=q$

Converting cm^2 to m^2 and plugging in values:

$(6000)*(8.86*10^{-12})*(4*10^{-4})=q$

$q=2.13*10^{-11}C$

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Example Question #311 : Electricity And Magnetism

A capacitor is placed in series with three parallel batteries. The plates are 1.5mm apart and each have a 10cm^2 surface area. Determine the charge on the positive plate.

Possible Answers:

$8.44*10^{-11}C$

None of these

$3.08*10^{-11}C$

$5.55*10^{-11}C$

$1.42*10^{-11}C$

Correct answer:

$1.42*10^{-11}C$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor. Since the batteries all have the same potential and are in parallel, the potential will be the same as if there was just one.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$

Combining equations:

V_1=E*d

Solving for E:

$\frac{V_1}{d}=E$

Converting to and plugging in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$

Combining equations:

$E*\epsilon_0*A=q$

Converting cm^2 to m^2 and plugging in values:

$(4000)*(8.86*10^{-12})*(4*10^{-4})=q$

$q=1.42*10^{-11}C$

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Example Question #1 : Resistors And Resistance

Suppose a circuit consists of a battery connected to four resistors, all of which are connected in parallel. If a fifth resistor is added in parallel to this circuit, how will the current in the other four resistors change?

Possible Answers:

The current through the four resistors will quadruple

There will be no change in the current through the four resistors

The current through the four resistors will double

There is not enough information given to answer the question

The current through the four resistors will halve

Correct answer:

There will be no change in the current through the four resistors

Explanation:

Since all the resistors are connected in parallel, they will all experience the same voltage drop, which is equal to the voltage of the battery. Furthermore, since each resistor has the same resistance, the current flowing through the four resistors will be equal.

The current flowing through each resistor can be shown by Ohm's law:

V=iR

Also, the total current of the circuit can be found by noting:

$V=i_{Total}R_{Total}$

Moreover, the total resistance can be calculated by noting that resistors in parallel add inversely.

$\frac{1}{R_{Total}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}$

Since each resistor is the same, we can note that:

$\frac{1}{R_{Total}}=\frac{4}{R}$

$R_{Total}=\frac{R}{4}$

Plugging $R_{Total}$ into the previous equation, we see that:

$i_{Total}=\frac{V}{\frac{R}{4}}=\frac{4V}{R}$

So in the case of four resistors connected in parallel, the total current in the circuit is equal to $\frac{4V}{R}$ . And since the overall current is split evenly between the four resistors, we see that:

$i=\frac{(\frac{4V}{R})}{4}=\frac{V}{R}$

Now, if we consider the case of five resistors connected in parallel, we will see that the total resistance is:

$\frac{1}{R_{Total}^{'}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}$

$\frac{1}{R_{Total}^{'}}=\frac{5}{R}$

$R_{Total}^{'}=\frac{R}{5}$

Therefore, the overall current in the circuit when a fifth resistor is added in parallel is:

$i_{Total}^{'}=\frac{V}{R_{Total}^{'}}=\frac{V}{(\frac{R}{5})}=\frac{5V}{R}$

And once again, since each of the five resistors are connected in parallel, the total current will be split evenly between them, thus giving:

$i^{'}=\frac{(\frac{5V}{R})}{5}=\frac{V}{R}$

$i=i^{'}=\frac{V}{R}$

Therefore, we see that the current through each resistor is the same, both before and after the addition of the fifth resistor.

To summarize, the voltage drop across each resistor is the same since they are connected in parallel. Also, since each resistor has the same resistance, the current through each is the same. Adding a fifth resistor in parallel decreased the overall resistance of the circuit and increased the overall current flowing through the circuit. But because this new, greater current is divided evenly among all the resistors, including the new one that was added, the current through each of the resistors does not change.

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Example Question #41 : Circuit Components

The following is a picture of a circuit with 4 resistors, labeled , , , and .

4resistorcircuit

What is the total resistance of the circuit if the values of the resistors are as follows?

$\begin{align*} A&=2\Omega \\ B&=2\Omega \\ C&=4\Omega \\ D&=3\Omega \end{align*}$

Possible Answers:

$11\Omega$

$10\Omega$

$1.2\Omega$

$5\Omega$

$6\Omega$

Correct answer:

$5\Omega$

Explanation:

Remember, the equations for find the total resistance of a circuit are as follows:

(Series) $R_{tot}&=R_1+R_2+...+R_n$

(Parallel) $\frac{1}{R_{tot}}&= \frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

Let's complete this in pieces. First, we see that and are in series, so they just add together. Next, we see that and are in parallel, so we invert the added inverses. Finally, we see that ABC and are in series, so we add them together.

$R_{AB}&=R_A+R_B$

$R_{AB}=2+2$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}=(\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}=(\frac{1}{2})^{-1}=2$

$R_{ABCD}&=R_{ABC}+R_D$

$R_{ABCD}=2+3$

$R_{ABCD}=5$

Therefore, the total resistance of the circuit is $5\Omega$ .

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Example Question #2 : Resistors And Resistance

Parallel circuit 1 jpeg

$R_1=5\Omega$

$R_2=7\Omega$

What is the total resistance of the circuit above?

Possible Answers:

$0.35\Omega$

$6.0\Omega$

$12\Omega$

$35\Omega$

$2.92\Omega$

Correct answer:

$2.92\Omega$

Explanation:

Use our rule for adding resistors in parallel.

$\frac{1}{R1}+\frac{1}{R2}=\frac{1}{R_{total}}$

Plug in known values.

$\frac{1}{5\Omega}+\frac{1}{7\Omega}=\frac{1}{R_{total}}$

$R_{total}=2.92\Omega$

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Example Question #2 : Resistors And Resistance

Photo 4 1

$R_A=3\Omega$

$R_B=6\Omega$

$R_C=9\Omega$

What is the total resistance of the circuit shown above?

Possible Answers:

$18\Omega$

None of these

$\frac{18}{11}\Omega$

$11\Omega$

$\frac{11}{18}\Omega$

Correct answer:

$\frac{18}{11}\Omega$

Explanation:

The equations for adding resistances in parallel, as in the diagram is:

$\frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

Solve.

$(R_{tot})^{-1}=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$(R_{tot})^{-1}= \left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)^{-1}$

$R_{tot}= \frac{18}{11}\Omega$

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Example Question #2 : Resistors And Resistance

Resistances:

$\begin{align*} A&= 1\Omega\\ B&= 2\Omega\\ C&= 3\Omega\\ D&= 4\Omega\\ E&= 5\Omega\\ F&=6\Omega \end{align*}$

If the circuit above is connected to a battery, what is the total power output of the system?

Possible Answers:

15.5000W

11.1375W

19.5467 W

16.1357 W

30.3030 W

Correct answer:

11.1375W

Explanation:

The rules for adding resistors are:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&= \frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

From the diagram, A, B, and C are in parallel, E and F are in parallel, and ABC , EF, and D are in series.

Find the resistance of A, B and C.

$R_{ABC}&=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$R_{ABC}=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\right)^{-1}$

$R_{ABC}= \frac{6}{11}\Omega$

Find the resistance of E, and F.

$R_{EF}=\left(\frac{1}{R_E}+\frac{1}{R_F}\right)^{-1}$

$R_{EF}= \left(\frac{1}{5}+\frac{1}{6}\right)^{-1}$

$R_{EF}= \frac{30}{11}\Omega$

Find the total equivalent resistance by using the rule for adding (systems of) resistors in series.

$R_{ABCDEF}&=R_{ABC}+R_D+R_{EF}$

$R_{ABCDEF}= \frac{6}{11}+4+\frac{30}{11}$

$R_{ABCDEF}= \frac{80}{11}\Omega$

Therefore the total resistance is $\frac{80}{11}\Omega$

The power equation is:

P=VI

Since we know total resistance and potential difference, we can find current. Use Ohm's law.

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

$I= \frac{9}{\frac{80}{11}}$

$I= \frac{99}{11}$

I=9A

Now we can find the power.

P=VI

$P= (9)(\frac{99}{80})$

$P= \frac{891}{80}W$

This fraction is equivalent to 11.1375 W

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Example Question #4 : Resistors And Resistance

Resistances:

$\begin{align*} A&= 3\Omega\\ B&= 1\Omega\\ C&= 2.4 \Omega\\ D&= 5\Omega\\ E&= 1.5\Omega\\ F&= 7\Omega \end{align*}$

What is the total resistance of the system in the diagram above?

Possible Answers:

$2.110\Omega$

$1.566\Omega$

$6.807\Omega$

$5.346\Omega$

$15.412\Omega$

Correct answer:

$6.807\Omega$

Explanation:

The rules for adding resistors are:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&= \frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

Note that these equations are the opposite of the equations used for adding capacitors.

From the diagram, A, B, and C are in parallel, E and F are in parallel, and ABC , EF, and D are in series.

Find the equivalent resistances for each system of resistors.

$R_{ABC}=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$R_{ABC}=\left(\frac{1}{3}+\frac{1}{1}+\frac{1}{2.4}\right)^{-1}$

$R_{ABC}= \frac{4}{7}\Omega$

Find the equivalent resistance of the other system of parallel resistors.

$R_{EF}&=\left(\frac{1}{R_E}+\frac{1}{R_F}\right)^{-1}$

$R_{EF}= \left(\frac{1}{1.5}+\frac{1}{7}\right)^{-1}$

$R_{EF}= \frac{21}{17}\Omega$

The three systems of resistors are in series. Find the equivalent resistance using the appropriate equation.

$R_{ABCDEF}=R_{ABC}+R_D+R_{EF}$

$R_{ABCDEF}= \frac{4}{7}+5+\frac{21}{17}$

$R_{ABCDEF}= \frac{810}{119}\Omega$

The total resistance is about $6.807\Omega$

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AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #9 : Capacitors And Electric Fields

Example Question #10 : Capacitors And Electric Fields

Example Question #31 : Circuit Components

Example Question #311 : Electricity And Magnetism

Example Question #1 : Resistors And Resistance

Example Question #41 : Circuit Components

Example Question #2 : Resistors And Resistance

Example Question #2 : Resistors And Resistance

Example Question #2 : Resistors And Resistance

Example Question #4 : Resistors And Resistance

All AP Physics 2 Resources

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