Two cells E_1 and E_2 having equal EMF E and | vedclass

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(A) The total $EMF$ of the series combination is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{total} = r_1 + r_2 + R$.The current

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$r_1 - r_2$

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(A) The total $EMF$ of the series combination is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{total} = r_1 + r_2 + R$.The current $I$ flowing through the circuit is given by $I = \frac{2E}{r_1 + r_2 + R}$.The potential difference across cell $E_1$ is given by $V_1 = E - Ir_1$.Given that $V_1 = 0$,we have $E - Ir_1 = 0$,which implies $E = Ir_1$.Substituting the value of $I$,we get $E = \left( \frac{2E}{r_1 + r_2 + R} \right) r_1$.Dividing both sides by $E$,we get $1 = \frac{2r_1}{r_1 + r_2 + R}$.Rearranging the terms,$r_1 + r_2 + R = 2r_1$.Therefore,$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

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