A torch battery consists of two cells of 1.45 | vedclass

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(B) The two cells are connected in series.The total electromotive force (emf) of the combination is $E_{total} = 2E = 2 	imes 1.45\,V = 2.9\,V$.The t

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$1.611$

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(B) The two cells are connected in series.The total electromotive force (emf) of the combination is $E_{total} = 2E = 2 	imes 1.45\,V = 2.9\,V$.The total internal resistance of the two cells is $r_{total} = 2r = 2 	imes 0.15\,\Omega = 0.3\,\Omega$.The external resistance of the lamp is $R = 1.5\,\Omega$.The total resistance of the circuit is $R_{total} = R + r_{total} = 1.5\,\Omega + 0.3\,\Omega = 1.8\,\Omega$.Using Ohm's law,the current $i$ is given by $i = \frac{E_{total}}{R_{total}} = \frac{2.9}{1.8} = \frac{29}{18} \approx 1.611\,A$.

$A$ torch battery consists of two cells of $1.45\,V$ and an internal resistance of $0.15\,\Omega$ each. Both cells send current through the filament of a lamp having a resistance of $1.5\,\Omega$. The value of the current will be ....... $A$.

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