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AP Physics 2 : Other Circuit Principles

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Example Questions

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:

$4\Omega$

$5\Omega$

$2\Omega$

$6\Omega$

$3\Omega$

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 #911 : Ap Physics 2

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:

$0.7\Omega$

$0.0068\Omega$

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

$0.068\Omega$

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

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 : Other Circuit Principles

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #101 : Circuits

Example Question #911 : Ap Physics 2

Example Question #3 : Other Circuit Principles

All AP Physics 2 Resources

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