In order to determine the e.m.f. of a storage | vedclass

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(A) Let $R$ be the resistance of the circuit. In the first case,the cells are connected in series aiding each other. The total $e.m.f.$ is $\varepsilo

Vedclass · Accepted Answer

$\varepsilon_1 = \frac{I_1 + I_2}{I_1 - I_2} \varepsilon_2$

Vedclass · Answer

(A) Let $R$ be the resistance of the circuit. In the first case,the cells are connected in series aiding each other. The total $e.m.f.$ is $\varepsilon_1 + \varepsilon_2$. Thus,$I_1 = \frac{\varepsilon_1 + \varepsilon_2}{R}$,which implies $\varepsilon_1 + \varepsilon_2 = I_1 R$. In the second case,the cells are connected in opposition. The total $e.m.f.$ is $\varepsilon_1 - \varepsilon_2$. Thus,$I_2 = \frac{\varepsilon_1 - \varepsilon_2}{R}$,which implies $\varepsilon_1 - \varepsilon_2 = I_2 R$. Dividing the two equations: $\frac{\varepsilon_1 + \varepsilon_2}{\varepsilon_1 - \varepsilon_2} = \frac{I_1 R}{I_2 R} = \frac{I_1}{I_2}$ $\varepsilon_1 I_2 + \varepsilon_2 I_2 = \varepsilon_1 I_1 - \varepsilon_2 I_1$ $\varepsilon_1 (I_1 - I_2) = \varepsilon_2 (I_1 + I_2)$ $\varepsilon_1 = \left( \frac{I_1 + I_2}{I_1 - I_2} \right) \varepsilon_2$

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