In the circuit shown in the figure,the resist | vedclass

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(B) The voltmeter is connected in parallel with the $3 \, k\Omega$ resistor. Let $R_v = 6 \, k\Omega$ be the resistance of the voltmeter and $R_2 = 3

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$5$

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(B) The voltmeter is connected in parallel with the $3 \, k\Omega$ resistor. Let $R_v = 6 \, k\Omega$ be the resistance of the voltmeter and $R_2 = 3 \, k\Omega$ be the resistor in parallel with it.The equivalent resistance $R_p$ of the parallel combination is given by:$R_p = \frac{R_v 	imes R_2}{R_v + R_2} = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2 \, k\Omega$Now,this $R_p$ is in series with the $2 \, k\Omega$ resistor. The total resistance of the circuit is $R_{total} = 2 \, k\Omega + 2 \, k\Omega = 4 \, k\Omega$.The total current in the circuit is $I = \frac{V_{source}}{R_{total}} = \frac{10 \, V}{4 \, k\Omega} = 2.5 \, mA$.The voltage across the parallel combination (which is the voltmeter reading) is:$V_{reading} = I 	imes R_p = 2.5 \, mA 	imes 2 \, k\Omega = 5 \, V$.

In the circuit shown in the figure,the resistance of the voltmeter is $6 \, k\Omega$. The voltmeter reading will be ................. $V$.

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