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(B) The voltmeter of resistance $R_v = 10 \, k \Omega = 10000 \, \Omega$ is connected in parallel with the $400 \, \Omega$ resistor. First,calculate t

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$1.95$

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(B) The voltmeter of resistance $R_v = 10 \, k \Omega = 10000 \, \Omega$ is connected in parallel with the $400 \, \Omega$ resistor. First,calculate the equivalent resistance $R_p$ of the parallel combination of the $400 \, \Omega$ resistor and the $10000 \, \Omega$ voltmeter: $R_p = \frac{400 	imes 10000}{400 + 10000} = \frac{4000000}{10400} \approx 384.6 \, \Omega \approx 385 \, \Omega$. Now,this combination is in series with the $800 \, \Omega$ resistor. The total resistance of the circuit is $R_{eq} = 385 \, \Omega + 800 \, \Omega = 1185 \, \Omega$. The potential difference across the parallel combination (measured by the voltmeter) is given by the voltage divider rule: $V = V_{total} 	imes \frac{R_p}{R_{eq}} = 6 \, V 	imes \frac{385 \, \Omega}{1185 \, \Omega} \approx 1.949 \, V$. Rounding this value,we get $1.95 \, V$.

Two resistors $400 \, \Omega$ and $800 \, \Omega$ are connected in series across a $6 \, V$ battery. The potential difference measured by a voltmeter of $10 \, k \Omega$ connected across the $400 \, \Omega$ resistor is close to $.... \, V$.

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