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AP Physics 2 : Circuit Properties

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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 : Circuit Properties

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:

$2\Omega$

$5\Omega$

$4\Omega$

$3\Omega$

$6\Omega$

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 #2 : Other Circuit Principles

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:

$0.068\Omega$

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

$0.7\Omega$

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

$0.0068\Omega$

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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AP Physics 2 : Circuit Properties

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 : Circuit Properties

Example Question #2 : Other Circuit Principles

Example Question #3 : Other Circuit Principles

All AP Physics 2 Resources

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