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SAT II Physics : Electricity

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Example Questions

Example Question #1 : Electric Circuits

If a circuit contains a 10V $\displaystyle 10V$ battery and a $2\Omega$ $\displaystyle 2\Omega$ resistor what is the current of the circuit?

Possible Answers:

$\frac{1}{5}A$ $\displaystyle \frac{1}{5}A$

20A $\displaystyle 20A$

100A $\displaystyle 100A$

40A $\displaystyle 40A$

$\displaystyle 5A$

Correct answer:

$\displaystyle 5A$

Explanation:

Ohm's law states:

V=IR $\displaystyle V=IR$

Where $\displaystyle V$ is voltage, $\displaystyle I$ is current, and $\displaystyle R$ is resistance.

We can substitute the given values from the question into the equation and solve:

10=I(2) $\displaystyle 10=I(2)$

$\frac{10}{2}=I$ $\displaystyle \frac{10}{2}=I$

5=I $\displaystyle 5=I$

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Example Question #41 : Electricity And Magnetism

There are three resistors in parallel in a circuit with resistances of $10\Omega$ $\displaystyle 10\Omega$, $20\Omega$ $\displaystyle 20\Omega$, and $30\Omega$ $\displaystyle 30\Omega$.

What is the equivalent resistance?

Possible Answers:

$1\Omega$ $\displaystyle 1\Omega$

$5.45\Omega$ $\displaystyle 5.45\Omega$

$0.017\Omega$ $\displaystyle 0.017\Omega$

$6000\Omega$ $\displaystyle 6000\Omega$

$60\Omega$ $\displaystyle 60\Omega$

Correct answer:

$5.45\Omega$ $\displaystyle 5.45\Omega$

Explanation:

The equation for resistors in parallel is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

We are given the values of the resistors. Using this formula, we can solve for the equivalent resistance.

Plug in the given values and solve.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

$\frac{1}{R_{eq}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{30\Omega}...$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{30\Omega}...$

$\frac{1}{R_{eq}}=\frac{11}{60\Omega}$ $\displaystyle \frac{1}{R_{eq}}=\frac{11}{60\Omega}$

$R_{eq}=\frac{60\Omega}{11}$ $\displaystyle R_{eq}=\frac{60\Omega}{11}$

$R_{eq}=5.45\Omega$ $\displaystyle R_{eq}=5.45\Omega$

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Example Question #1 : Resistors

A voltage of $\displaystyle 6V$ applied between the ends of a wire results in a current of 60A $\displaystyle 60A$. What is the resistance of the wire?

Possible Answers:

$0.2\Omega$ $\displaystyle 0.2\Omega$

$1\Omega$ $\displaystyle 1\Omega$

$5\Omega$ $\displaystyle 5\Omega$

$10\Omega$ $\displaystyle 10\Omega$

$0.1\Omega$ $\displaystyle 0.1\Omega$

Correct answer:

$0.1\Omega$ $\displaystyle 0.1\Omega$

Explanation:

This question can be solved using Ohm's law:

V=IR $\displaystyle V=IR$

We are given the voltage and the current. Using these values, we can solve for the resistance.

$\displaystyle 6V=(60A)R$

$R=\frac{6V}{60A}=0.1\Omega$ $\displaystyle R=\frac{6V}{60A}=0.1\Omega$

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Example Question #2 : Electricity And Magnetism

A capacitor with capacitance $\displaystyle C$ is constructed by putting a thin piece of cardboard between two copper plates, then each plate is connected to a battery. If the copper plates are cut in half then what is the effect on the capacitance of the circuit?

Possible Answers:

$C^{2}$ $\displaystyle C^{2}$

$\frac{1}{C}$ $\displaystyle \frac{1}{C}$

$\frac{1}{2}C$ $\displaystyle \frac{1}{2}C$

$\displaystyle 2C$

$\displaystyle C$

Correct answer:

$\frac{1}{2}C$ $\displaystyle \frac{1}{2}C$

Explanation:

Capacitance of a circuit is defined by the equation:

$C=\varepsilon\frac{A}{d}$ $\displaystyle C=\varepsilon\frac{A}{d}$

Where $\displaystyle C$ is the capacitance, $\varepsilon$ $\displaystyle \varepsilon$ is a constant of nature, $\displaystyle A$ is the area of the capacitor, and $\displaystyle d$ is the distance between the two plates

Since the metal plates are cut in half, the area is halved. We can substitute $\frac{1}{2}A$ $\displaystyle \frac{1}{2}A$ in for $\displaystyle A$

The result is that the capacitance is half the original quantity.

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Example Question #1 : Coulomb's Law

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Possible Answers:

quadruple

be quartered

double

remain unchanged

be halved

Correct answer:

be quartered

Explanation:

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

$F_e=k\frac{q_1q_2}{r^2}$ $\displaystyle F_e=k\frac{q_1q_2}{r^2}$

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$ $\displaystyle F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

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Example Question #1 : Electric Potential Energy

On which of the following does the amount of work required to move a charge in an electric field depend?

Possible Answers:

Only the magnetic field

Only the path traveled

Neither the potential nor the path traveled

Both the potential and the path traveled

Only the change in potential

Correct answer:

Only the change in potential

Explanation:

Work done by electric field is defined:

$W=-\Delta PE$ $\displaystyle W=-\Delta PE$

Notice that the only variable in the equation is the potential, so this is the only quantity on which work depends.

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SAT II Physics : Electricity

Study concepts, example questions & explanations for SAT II Physics

All SAT II Physics Resources

Example Questions

Example Question #1 : Electric Circuits

Example Question #41 : Electricity And Magnetism

Example Question #1 : Resistors

Example Question #2 : Electricity And Magnetism

Example Question #1 : Coulomb's Law

Example Question #1 : Electric Potential Energy

All SAT II Physics Resources

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