In the circuit shown, a four-wire potentiomet | vedclass

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(C) The total $EMF$ of the two cells in series is $E_{eq} = 1.5\, V + 1.5\, V = 3.0\, V$.The total internal resistance of the two cells is $r_{int} =

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$0.25$

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(C) The total $EMF$ of the two cells in series is $E_{eq} = 1.5\, V + 1.5\, V = 3.0\, V$.The total internal resistance of the two cells is $r_{int} = 0.5\, \Omega + 0.5\, \Omega = 1.0\, \Omega$.The total resistance of the circuit is $R_{total} = R_{ext} + r_{int} + R_{wire} = 1.0\, \Omega + 1.0\, \Omega + (400\, cm 	imes 0.01\, \Omega/cm) = 2.0\, \Omega + 4.0\, \Omega = 6.0\, \Omega$.The current in the circuit is $i = \frac{E_{eq}}{R_{total}} = \frac{3.0\, V}{6.0\, \Omega} = 0.5\, A$.The voltmeter measures the potential difference across the $50\, cm$ length of the wire.The resistance of this $50\, cm$ segment is $R_{50} = 50\, cm 	imes 0.01\, \Omega/cm = 0.5\, \Omega$.Therefore, the voltmeter reading is $V = i 	imes R_{50} = 0.5\, A 	imes 0.5\, \Omega = 0.25\, V$.

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