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(D) The voltmeter has a resistance of $1000 \, \Omega$. It is connected in parallel with a $500 \, \Omega$ resistor. The equivalent resistance $R_{AB}

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$4$

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(D) The voltmeter has a resistance of $1000 \, \Omega$. It is connected in parallel with a $500 \, \Omega$ resistor. The equivalent resistance $R_{AB}$ between points $A$ and $B$ is given by: $R_{AB} = \frac{1000 	imes 500}{1000 + 500} = \frac{500000}{1500} = \frac{1000}{3} \, \Omega$. This combination is in series with another $500 \, \Omega$ resistor. The total equivalent resistance of the circuit is: $R_{eq} = R_{AB} + 500 = \frac{1000}{3} + 500 = \frac{1000 + 1500}{3} = \frac{2500}{3} \, \Omega$. The total current $i$ drawn from the $10 \, V$ battery is: $i = \frac{V_{total}}{R_{eq}} = \frac{10}{2500/3} = \frac{30}{2500} = \frac{3}{250} \, A$. The reading of the voltmeter is the potential difference across $AB$: $V_{AB} = i 	imes R_{AB} = \frac{3}{250} 	imes \frac{1000}{3} = 4 \, V$.

What is the reading of the voltmeter in the following figure (in $V$)?

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