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(A) When the voltmeter of resistance $R_v = 2000 \,\Omega$ is connected in parallel with the $500 \,\Omega$ resistor,the equivalent resistance $R_p$ o

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$8$

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(A) When the voltmeter of resistance $R_v = 2000 \,\Omega$ is connected in parallel with the $500 \,\Omega$ resistor,the equivalent resistance $R_p$ of this parallel combination is:$R_p = \frac{500 	imes 2000}{500 + 2000} = \frac{1000000}{2500} = 400 \,\Omega$Now,the circuit consists of $R_p = 400 \,\Omega$ and $R_2 = 600 \,\Omega$ in series with a $20 \,V$ battery.The total resistance of the circuit is $R_{eq} = 400 + 600 = 1000 \,\Omega$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{20}{1000} = 0.02 \,A$.The voltmeter reading is the potential difference across the parallel combination $R_p$:$V_{reading} = I 	imes R_p = 0.02 	imes 400 = 8 \,V$.

Two resistors are connected in series across a battery as shown in the figure. If a voltmeter of resistance $2000 \,\Omega$ is used to measure the potential difference across the $500 \,\Omega$ resistor,the reading of the voltmeter will be ............... $V$.

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