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Question 25 of 36

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The figure presents an electrical circuit with a battery in series with a resistor. The battery is on the left side of the circuit and is labeled with the emf symbol (the Greek letter epsilon). The resistor is on the right side of the circuit and is labeled R. On the top side of the circuit there is an arrow pointing toward the right that is labeled I.

A single resistor $R$ is connected to a battery as shown above. The current is $I$ and the power dissipated as heat is $P$ . The circuit is changed by doubling the emf $ε$ of the battery while $R$ is kept constant.

After the change, the power dissipated in $R$ is

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$P over 4$

Correct Answer:

No

$P over 2$

Correct Answer:

No

$P$

Correct Answer:

No

$2 P$

Correct Answer:

No

$4 P$

Correct Answer:

Yes

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The correct answer is $4 P$ . The power $P$ dissipated by a resistor is given by $P = I V$ , where $I$ is the current in the resistor and $V$ is the voltage across it. When used in combination with Ohm’s law, the power can also be written as $P = V^{2} / R$ , where $R$ is the resistance. If the emf is doubled while $R$ remains the same, $P$ is quadrupled.

Question Difficulty:

hard

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Subject Test Physics

Question 25 of 36

Learn more about plans to discontinue SAT Subject Tests

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