Two cells of emfs 1 V and 2 V and internal re | vedclass

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(C) For cells in parallel, the equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by:$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r

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$3$

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(C) For cells in parallel, the equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by:$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$ and $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$.Given $E_1 = 1$ $V$, $r_1 = 2 \Omega$, $E_2 = 2$ $V$, $r_2 = 1 \Omega$.$E_{eq} = \frac{1/2 + 2/1}{1/2 + 1/1} = \frac{2.5}{1.5} = \frac{5}{3}$ $V$.$r_{eq} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3} \Omega$.Given current $I = 1$ $A$ through external resistance $R$, $I = \frac{E_{eq}}{R + r_{eq}} \Rightarrow 1 = \frac{5/3}{R + 2/3}$.$R + 2/3 = 5/3 \Rightarrow R = 1 \Omega$.If the polarity of one cell is reversed, the new equivalent emf $E'_{eq} = \frac{E_1/r_1 - E_2/r_2}{1/r_1 + 1/r_2} = \frac{0.5 - 2}{1.5} = \frac{-1.5}{1.5} = -1$ $V$.The magnitude of the new current $I' = \frac{|E'_{eq}|}{R + r_{eq}} = \frac{1}{1 + 2/3} = \frac{1}{5/3} = \frac{3}{5}$ $A$.Comparing $I' = \frac{3}{5}$ $A$ with $\frac{\alpha}{5}$ $A$, we get $\alpha = 3$.

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