A voltmeter of resistance 1000,Omega is conne | vedclass

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(D) The voltmeter is connected in parallel with the $500\,\Omega$ resistor. The equivalent resistance $R_p$ of this parallel combination is given by:$

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$4$

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(D) The voltmeter is connected in parallel with the $500\,\Omega$ resistor. The equivalent resistance $R_p$ of this parallel combination is given by:$R_p = \frac{1000 	imes 500}{1000 + 500} = \frac{500000}{1500} = \frac{1000}{3}\,\Omega$This parallel combination is in series with the other $500\,\Omega$ resistor. The total resistance $R_{eq}$ of the circuit is:$R_{eq} = \frac{1000}{3} + 500 = \frac{1000 + 1500}{3} = \frac{2500}{3}\,\Omega$The total current $I$ flowing from the $10\,V$ battery is:$I = \frac{V}{R_{eq}} = \frac{10}{2500/3} = \frac{30}{2500} = \frac{3}{250}\,A$The reading of the voltmeter is the potential difference across the parallel combination,which is $V_v = I 	imes R_p$:$V_v = \frac{3}{250} 	imes \frac{1000}{3} = 4\,V$

$A$ voltmeter of resistance $1000\,\Omega$ is connected across a resistance of $500\,\Omega$ in the given circuit. What will be the reading of the voltmeter in $V$?

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