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(C) Let the two resistors be $R_1 = 100 \Omega$ and $R_2 = 100 \Omega$. The voltmeter with resistance $R_v = 400 \Omega$ is connected in parallel with

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$4$

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(C) Let the two resistors be $R_1 = 100 \Omega$ and $R_2 = 100 \Omega$. The voltmeter with resistance $R_v = 400 \Omega$ is connected in parallel with $R_2$.The equivalent resistance of the parallel combination of $R_2$ and $R_v$ is:$R_p = \frac{R_2 	imes R_v}{R_2 + R_v} = \frac{100 	imes 400}{100 + 400} = \frac{40000}{500} = 80 \Omega$The total resistance of the circuit is $R_{eq} = R_1 + R_p = 100 + 80 = 180 \Omega$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{9}{180} = \frac{1}{20} 	ext{ A} = 0.05 	ext{ A}$.The voltage drop across the parallel combination (which is the voltmeter reading) is $V_v = I 	imes R_p = 0.05 	imes 80 = 4 	ext{ V}$.

Two resistors of $100 \Omega$ each are connected in series with a $9 \text{ V}$ battery. $A$ voltmeter of $400 \Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is . . . . . . $V$.

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