A 10,V battery with internal resistance 1,Ome | vedclass

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(C) Since the two cells are connected such that they oppose each other,the effective $EMF$ in the closed circuit is $E_{eff} = 15 - 10 = 5\,V$.The tot

Vedclass · Accepted Answer

$13.1$

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(C) Since the two cells are connected such that they oppose each other,the effective $EMF$ in the closed circuit is $E_{eff} = 15 - 10 = 5\,V$.The total resistance of the circuit is $R_{total} = 1 + 0.6 = 1.6\,\Omega$ (as the internal resistances are in series in the closed loop).The current flowing in the circuit is given by $I = \frac{E_{eff}}{R_{total}} = \frac{5}{1.6} = 3.125\,A$.The voltmeter measures the terminal potential difference across the batteries. For the $15\,V$ battery (which is discharging),the terminal voltage is $V = E_1 - I r_1 = 15 - (3.125 	imes 0.6) = 15 - 1.875 = 13.125\,V$.Alternatively,for the $10\,V$ battery (which is charging),the terminal voltage is $V = E_2 + I r_2 = 10 + (3.125 	imes 1) = 10 + 3.125 = 13.125\,V$.Thus,the reading in the voltmeter is approximately $13.1\,V$.

$A$ $10\,V$ battery with internal resistance $1\,\Omega$ and a $15\,V$ battery with internal resistance $0.6\,\Omega$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to ................ $V$.

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