Two cells E_1 and E_2 having equal e.m.f 'E' | vedclass

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Question

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

Vedclass · Accepted Answer

$r_1 - r_2$

Vedclass · Answer

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

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