Voltmeters V_1 and V_2 are connected in serie | vedclass

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(D) The resistance of voltmeter $V_1$ is $R_1 = 80 \, V 	imes 200 \, \Omega/V = 16000 \, \Omega = 16 \, k\Omega$.Since the voltmeters are connected i

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$240$

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(D) The resistance of voltmeter $V_1$ is $R_1 = 80 \, V 	imes 200 \, \Omega/V = 16000 \, \Omega = 16 \, k\Omega$.Since the voltmeters are connected in series, the current $I$ flowing through both is the same.$I = \frac{V_1}{R_1} = \frac{80 \, V}{16 	imes 10^3 \, \Omega} = 5 	imes 10^{-3} \, A$.The potential difference across $V_2$ is $V_2 = I 	imes R_2 = (5 	imes 10^{-3} \, A) 	imes (32 	imes 10^3 \, \Omega) = 160 \, V$.The total line voltage $V$ is the sum of the potential differences across the series components:$V = V_1 + V_2 = 80 \, V + 160 \, V = 240 \, V$.

Voltmeters $V_1$ and $V_2$ are connected in series across a $D.C.$ line. $V_1$ reads $80 \, V$ and has a resistance per volt of $200 \, \Omega/V$. $V_2$ has a total resistance of $32 \, k\Omega$. The line voltage is ............. (in $V$)

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