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

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$1.33$

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(C) The voltmeter is connected in parallel with the $80 \, \Omega$ resistor. First,calculate the equivalent resistance of the parallel combination of the voltmeter $(80 \, \Omega)$ and the resistor $(80 \, \Omega)$: $R_p = \frac{80 	imes 80}{80 + 80} = \frac{6400}{160} = 40 \, \Omega$. This parallel combination is in series with the $20 \, \Omega$ resistor. Total resistance of the circuit $R_{eq} = 40 \, \Omega + 20 \, \Omega = 60 \, \Omega$. The main current $i$ flowing from the cell is $i = \frac{V}{R_{eq}} = \frac{2 \, V}{60 \, \Omega} = \frac{1}{30} \, A$. The reading of the voltmeter is the potential difference across the parallel combination: $V_{reading} = i 	imes R_p = \left( \frac{1}{30} \, A ight) 	imes 40 \, \Omega = \frac{4}{3} \, V \approx 1.33 \, V$.

In the adjoining circuit,the $e.m.f.$ of the cell is $2 \, V$ and the internal resistance is negligible. The resistance of the voltmeter is $80 \, \Omega$. The reading of the voltmeter will be ............. $V$.

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