Resistances R_1 and R_2,each 60,Omega,are con | vedclass

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(A) The voltmeter is connected in parallel with resistor $R_2$. The resistance of the voltmeter is $R_v = 120\,\Omega$ and $R_2 = 60\,\Omega$.The equi

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$48\,V$

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(A) The voltmeter is connected in parallel with resistor $R_2$. The resistance of the voltmeter is $R_v = 120\,\Omega$ and $R_2 = 60\,\Omega$.The equivalent resistance $R_p$ of the parallel combination of $R_2$ and $R_v$ is given by:$R_p = \frac{R_2 \cdot R_v}{R_2 + R_v} = \frac{60 	imes 120}{60 + 120} = \frac{7200}{180} = 40\,\Omega$.Now,the circuit consists of $R_1$ in series with the parallel combination $R_p$. The total resistance of the circuit is:$R_{total} = R_1 + R_p = 60 + 40 = 100\,\Omega$.The total current $I$ flowing through the circuit is:$I = \frac{V}{R_{total}} = \frac{120}{100} = 1.2\,A$.The reading of the voltmeter is the potential difference across the parallel combination $R_p$:$V_{voltmeter} = I 	imes R_p = 1.2 	imes 40 = 48\,V$.

Resistances $R_1$ and $R_2$,each $60\,\Omega$,are connected in series as shown in the figure. The potential difference between $A$ and $B$ is maintained at $120\,V$. What will be the reading of the voltmeter connected between points $C$ and $D$ if the resistance of the voltmeter is $120\,\Omega$?

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