Two cells of emf 1 V and 2 V and internal r | vedclass

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(C) Case $1$: Series connection.Equivalent emf,$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2 = 1 \ V + 2 \ V = 3 \ V$.Equivalent internal resistan

Vedclass · Accepted Answer

$4$

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(C) Case $1$: Series connection.Equivalent emf,$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2 = 1 \ V + 2 \ V = 3 \ V$.Equivalent internal resistance,$r_{eq} = r_1 + r_2 = 2 \ \Omega + 1 \ \Omega = 3 \ \Omega$.Total resistance,$R_{total} = R + r_{eq} = 6 \ \Omega + 3 \ \Omega = 9 \ \Omega$.Current $I_1 = \frac{\varepsilon_{eq}}{R_{total}} = \frac{3 \ V}{9 \ \Omega} = \frac{1}{3} \ A$.Case $2$: Parallel connection.Equivalent emf,$\varepsilon_{eq} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{2.5}{1.5} = \frac{5}{3} \ V$.Equivalent internal resistance,$r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3} \ \Omega$.Total resistance,$R_{total} = R + r_{eq} = 6 \ \Omega + \frac{2}{3} \ \Omega = \frac{20}{3} \ \Omega$.Current $I_2 = \frac{\varepsilon_{eq}}{R_{total}} = \frac{5/3}{20/3} = \frac{5}{20} = \frac{1}{4} \ A$.Ratio $\frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac{4}{3}$.Given $\frac{I_1}{I_2} = \frac{x}{3}$,therefore $x = 4$.

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