In the circuits shown below, the readings of | vedclass

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(C) In Circuit $1$, the voltmeter $V_{1}$ is connected in parallel with the $10 \; \Omega$ resistor. Since an ideal voltmeter has infinite resistance,

Vedclass · Accepted Answer

$V_{1} = V_{2}$ and $I_{1} = I_{2}$

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(C) In Circuit $1$, the voltmeter $V_{1}$ is connected in parallel with the $10 \; \Omega$ resistor. Since an ideal voltmeter has infinite resistance, it draws no current. Thus, the voltage across the $10 \; \Omega$ resistor is $10 \; V$, so $V_{1} = 10 \; V$. The current $I_{1}$ is given by $I_{1} = \frac{10 \; V}{10 \; \Omega} = 1 \; A$.In Circuit $2$, the voltmeter $V_{2}$ is connected in series with a $10 \; \Omega$ resistor, and this combination is in parallel with the main $10 \; \Omega$ resistor. An ideal voltmeter has infinite resistance, so the branch containing the voltmeter and the $10 \; \Omega$ resistor draws no current. The voltage across this branch is still $10 \; V$. Since the voltmeter is ideal, the entire potential drop of $10 \; V$ appears across the voltmeter itself, so $V_{2} = 10 \; V$. The current $I_{2}$ flows only through the main $10 \; \Omega$ resistor, so $I_{2} = \frac{10 \; V}{10 \; \Omega} = 1 \; A$.Therefore, $V_{1} = V_{2} = 10 \; V$ and $I_{1} = I_{2} = 1 \; A$.

In the circuits shown below, the readings of the voltmeters and the ammeters will be:

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