JEE Main 2025 Apr 3 Shift 2, Physics Q23: Kirchhoff’s Laws

JEE Main2025Apr 3, Shift 2Physics

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 ______.

Your answer

Solution

Series case: net emf $= 1 + 2 = 3$ V, total resistance $= 2 + 1 + 6 = 9 Ω$ .

$I_{1} = \frac{3}{9} = \frac{1}{3}$ A.

Parallel case (cells in parallel, then external $6 Ω$ ). Equivalent emf:

$ε_{e q} = \frac{ε _{1} / r _{1} + ε _{2} / r _{2}}{1/ r _{1} + 1/ r _{2}} = \frac{1/2 + 2/1}{1/2 + 1} = \frac{5/2}{3/2} = \frac{5}{3}$ V.

Equivalent internal resistance:

$r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}} = \frac{2}{3} Ω$ .

Total resistance $= \frac{2}{3} + 6 = \frac{20}{3} Ω$ .

$I_{2} = \frac{5/3}{20/3} = \frac{5}{20} = \frac{1}{4}$ A.

$\frac{I _{1}}{I _{2}} = \frac{1/3}{1/4} = \frac{4}{3}$ , so $x = 4$ .

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