Two identical cells each of emf 1.5,V are con | vedclass

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(A) The total emf of two cells connected in series is $E_{eq} = 1.5\,V + 1.5\,V = 3.0\,V$.Let $r$ be the internal resistance of each cell. The total i

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$5$

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(A) The total emf of two cells connected in series is $E_{eq} = 1.5\,V + 1.5\,V = 3.0\,V$.Let $r$ be the internal resistance of each cell. The total internal resistance in the series circuit is $R_{int} = r + r = 2r$.The total resistance of the circuit is $R_{total} = R + R_{int} = 10 + 2r$.The current $I$ flowing through the circuit is given by $I = \frac{E_{eq}}{R_{total}} = \frac{3}{10 + 2r}$.The voltage across the $10\,\Omega$ resistor is given as $V = 1.5\,V$.Using Ohm's law,$V = I 	imes R$,we have $1.5 = \left( \frac{3}{10 + 2r} ight) 	imes 10$.$1.5(10 + 2r) = 30$.$15 + 3r = 30$.$3r = 15$.$r = 5\,\Omega$.

Two identical cells each of emf $1.5\,V$ are connected in series across a $10\,\Omega$ resistance. An ideal voltmeter connected across $10\,\Omega$ resistance reads $1.5\,V$. The internal resistance of each cell is $......\Omega$.

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