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AP Physics 2 : Circuits

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #25 : Circuit Power

A lone $3\Omega$ resistor is placed in series with a battery. How will adding a second $3\Omega$ resistor in series affect the power output of the circuit?

Possible Answers:

Cut it in half

Double it

None of these

It will be unaffected

Quadruple it

Correct answer:

Cut it in half

Explanation:

Definition of electrical power:

P=IV

Ohm's law:

V=IR

Combining equations:

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

When adding resistors in series, they add directly

$R_1+R_2+.....=R_{total}$

Thus, in this case, resistance would increase, and if the voltage is kept constant, the power dissipated by the circuit would decrease. Doubling the resistance would cut the power output in half.

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Example Question #24 : Circuit Power

A lone $3\Omega$ resistor is placed in series with a battery. How will adding a second $3\Omega$ resistor in parallel affect the power output of the circuit?

Possible Answers:

Quadruple it

Double it

Cut it in half

None of these

It would stay the same

Correct answer:

Double it

Explanation:

Definition of electrical power:

P=IV

Ohm's law:

V=IR

Combining equations:

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

When adding resistors in parallel, they add inversely

$\frac{1}{R_1}+\frac{1}{R_2}+.....=\frac{1}{R_{total}}$

Thus, in this case, resistance would decrease, and if the voltage is kept constant, the power dissipated by the circuit would increase. Adding identical resistors in parallel would halve the resistance, which would double the power dissipated.

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Example Question #101 : Circuits

You have 4 resistors and an ammeter arranged as shown in the diagram below.

An ammeter measures current. The ammeter in this setup reads 0A. What is the resistance of ?

Possible Answers:

$3\Omega$

$2\Omega$

$4\Omega$

$6\Omega$

$5\Omega$

Correct answer:

$6\Omega$

Explanation:

This setup is called a Wheatstone bridge. It's used to find the resistance of a resistor with an unknown value. When the ammeter reads 0, the two sides are at equipotential, so

$\frac{1}{3}=\frac{2}{R}$ .

Therefore, the resistance of R is $6\Omega$ .

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Example Question #102 : Circuits

What is the resistance of a $\textup {1.5m}$ long copper wire with a diameter of $\textup {2.2mm}$ ?

$\rho_{copper}=1.72 * 10^{-8}\Omega\textup{m}$

Possible Answers:

$0.7\Omega$

$6.78 * 10^{-6}\Omega$

$6.78 * 10^{-9}\Omega$

$0.0068\Omega$

$0.068\Omega$

Correct answer:

$0.0068\Omega$

Explanation:

Write the following formula to find the resistance of the copper wire.

$R=\rho(\frac{L}{A})$

where $\rho$ is the resistivity in $\Omega \textup{m}$ , $\textup L$ is the length of the wire in $\textup m$ , and $\textup A$ is the cross sectional area of the wire in $\textup{m}^\textup {2}$ .

The resistivity of copper is: $1.72 * 10^{-8} \Omega \textup{m}$ .

$\textup L=1.5 \textup{m}$

$\textup A=\pi r^2=\pi(0.0011m)^2 = 1.21*10^{-6} \pi m^2$

Substitute the givens to the resistance formula and solve.

$\textup R=(1.72 * 10^{-8}\Omega \textup{m})(\frac{1.5 \textup{m}}{1.21*10^-6\pi \textup{m}^2})$

$\textup R= 0.0068\Omega$

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Example Question #3 : Other Circuit Principles

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the total resistance of the given circuit.

Possible Answers:

$3\Omega$

$3.67\Omega$

$12\Omega$

$2.62\Omega$

Correct answer:

$2.62\Omega$

Explanation:

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

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Example Question #1 : Capacitors And Capacitance

Consider the circuit:

Circuit_3__capacitors_

What is the total equivalent capacitance?

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Possible Answers:

3.08F

2.77F

4.01F

9.25F

11.31F

Correct answer:

3.08F

Explanation:

Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:

$C_{eq}=C2+C3 = 7F+1F= 8F$

The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:

$\frac{1}{C_{tot}}=\sum\frac{1}{C} = \frac{1}{5}+\frac{1}{8} = \frac{13}{40}$

$C_{tot} = 3.08F$

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Example Question #1 : Capacitors And Capacitance

You have 3 capacitors in series. Their capcitance's are $4\mu F$ , $3\mu F$ , and $2\mu F$ . What is the total capacitance of the system?

Possible Answers:

$\frac{13}{12} \mu F$

$\frac{12}{13} \mu F$

None of the other answers is correct

$9 \mu F$

$\frac{1}{9} \mu F$

Correct answer:

$\frac{12}{13} \mu F$

Explanation:

To find the total capacitance of capacitors in series, you use the following equation:

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...+\frac{1}{C_n}$ .

Our values for C_1 , C_2 , and C_3 are 4, 3, and 2. Now, we can plug in our values to find the answer.

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C_{tot}}=\frac{1}{4}+\frac{1}{3}+\frac{1}{2}$

$\frac{1}{C_{tot}}= \frac{13}{12}$

The answer we have is the inverse of this. Therefore, the total capacitance is $\frac{12}{13} \mu F$ .

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Example Question #1 : Circuit Components

You have 4 capacitors, , , , and , arranged as shown in the diagram below.

4capacitorcircuit

Their capacitances are as follows:

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

What is the total capacitance of the circuit?

Possible Answers:

$\frac{36}{49}F$

$\frac{49}{36}F$

Correct answer:

Explanation:

Remember, the equations for adding capacitances are as follows:

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

Capacitors and are in series, and are in parallel, and ABC and are in parallel.

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{3}+\frac{1}{6})^{-1}$

$C_{AB}= 2$

$C_{ABC}=C_{AB}+C_C$

$C_{ABC}=2+4$

$C_{ABC}= 6$

$C_{total}=C_{ABC}+C_{D}$

$C_{total}=6+1$

$C_{total}=7$

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Example Question #2 : Capacitors And Capacitance

Physics2set1q10

In this parallel-plate capacitor, the distance between the plates is 3cm , the area of each plate is 15cm^2 , and the voltage across them is .

Calculate the capacitance.

$\epsilon_0=8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}$

Possible Answers:

$8.32\cdot 10^{-13}F$

$1.29\cdot 10^{-12}F$

$4.43\cdot 10^{-13}F$

$3.33\cdot 10^{-13}F$

$9.31\cdot 10^{-11}F$

Correct answer:

$4.43\cdot 10^{-13}F$

Explanation:

The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.

For a parallel-plate capacitor,

$C=\frac{\epsilon _{0}A}{d}$

Plugging in the numbers given results in

$C=\frac{8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}\cdot0.0015m^2}{0.03m}$

$C=4.43\cdot 10^{-13}F$

Be careful to convert the units to meters!

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Example Question #1 : Capacitors And Capacitance

Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?

Possible Answers:

The total capacitance is the arithmetic mean of the individual capacitances.

The total capacitance is greater than any of the individual capacitances.

Nothing can be said for certain about the total capacitance.

The total capacitance is less than any of the individual capacitances.

The total capacitance is somewhere between the highest capacitance and the lowest capacitance (inclusive).

Correct answer:

The total capacitance is greater than any of the individual capacitances.

Explanation:

Capacitors in parallel combine according to the following equation:

$C_{tot}=C_1+C_2+...+C_n$ .

Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.

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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 #25 : Circuit Power

Example Question #24 : Circuit Power

Example Question #101 : Circuits

Example Question #102 : Circuits

Example Question #3 : Other Circuit Principles

Example Question #1 : Capacitors And Capacitance

Example Question #1 : Capacitors And Capacitance

Example Question #1 : Circuit Components

Example Question #2 : Capacitors And Capacitance

Example Question #1 : Capacitors And Capacitance

All AP Physics 2 Resources

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