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(A) Let the two resistors be $R_1 = 100\,\Omega$ and $R_2 = 100\,\Omega$. The voltmeter is connected in parallel with $R_1$.The equivalent resistance

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Let the two resistors be $R_1 = 100\,\Omega$ and $R_2 = 100\,\Omega$. The voltmeter is connected in parallel with $R_1$.The equivalent resistance of the parallel combination of $R_1$ and the voltmeter $(R_v = 400\,\Omega)$ is:$R_p = \frac{R_1 	imes R_v}{R_1 + R_v} = \frac{100 	imes 400}{100 + 400} = \frac{40000}{500} = 80\,\Omega$.The total resistance of the circuit is $R_{eq} = R_p + R_2 = 80 + 100 = 180\,\Omega$.The total current drawn from the cell is $I = \frac{V}{R_{eq}} = \frac{90}{180} = 0.5\,A$.The voltmeter measures the potential difference across the parallel combination $R_p$,which is:$V_{reading} = I 	imes R_p = 0.5 	imes 80 = 40\,V$.

$A$ cell of emf $90\,V$ is connected across a series combination of two resistors,each of $100\,\Omega$ resistance. $A$ voltmeter of resistance $400\,\Omega$ is used to measure the potential difference across one of the resistors. The reading of the voltmeter will be $.........\,V$.

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