In the adjoining figure,the emf of the cell i | vedclass

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(B) The voltmeter is connected in parallel with the $80 \, \Omega$ resistor.Let the equivalent resistance of the parallel combination of the voltmeter

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(B) The voltmeter is connected in parallel with the $80 \, \Omega$ resistor.Let the equivalent resistance of the parallel combination of the voltmeter $(80 \, \Omega)$ and the resistor $(80 \, \Omega)$ be $R'$.$R' = \frac{80 	imes 80}{80 + 80} = \frac{6400}{160} = 40 \, \Omega$.Now,this $R'$ is in series with the $20 \, \Omega$ resistor.So,the total equivalent resistance of the circuit is $R_{eq} = 20 + 40 = 60 \, \Omega$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{2}{60} = \frac{1}{30} \, A$.The reading of the voltmeter is the potential difference across the parallel combination $R'$,which is given by $V' = I 	imes R'$.$V' = \frac{1}{30} 	imes 40 = \frac{4}{3} = 1.33 \, V$.

In the adjoining figure,the $emf$ of the cell is $2 \, V$ and its internal resistance is negligible. The resistance of the voltmeter is $80 \, \Omega$. The reading of the voltmeter will be ................ $volt$.

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