Two cells of emf 1 V and 2 V and internal resistance 2 and 1 , respectively, are connected in series with an external resistance of 6 . The total current in the circuit is . Now the same two cells in parallel configuration are connected to same external

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Solution

Q.
Two cells of emf 1 V and 2 V and internal resistance 2 $Ω$ and 1 $Ω$ , respectively, are connected in series with an external resistance of 6 $Ω$ . The total current in the circuit is $I_{1}$ . Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $I_{2}$ . The value of $(\frac{I_{1}}{I_{2}})$ is $\frac{X}{3}$ . The value of X is _________. [2025]

Ans.
(4)

$I_{1} = \frac{ε_{1} + ε_{2}}{r_{1} + r_{2} + R} = \frac{1 + 2}{2 + 1 + 6} = \frac{3}{9} = \frac{1}{3} A$

$ε_{eq} = \frac{\frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{5}{3} V$

$r_{eq} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} + R = \frac{2 \times 1}{3} + 6 = \frac{20}{3} Ω$

$I_{2} = \frac{ε_{eq}}{r_{eq}} = \frac{\frac{5}{3}}{\frac{20}{3}} = \frac{1}{4} A$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{3} \times \frac{4}{1} = \frac{4}{3}$

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