JEE Main 2025 — Current Electricity Question with Solution

Series Connection:

Total emf = 1 V + 2 V = 3 V.

Total resistance = 2 Ω + 1 Ω + 6 Ω = 9 Ω.

Current $$I_1=\frac{\text{emf}}{\text{resistance}}=\frac{3}{9}=\frac{1}{3}\text{ A}.$$

Parallel Connection:

Equivalent emf: $${\epsilon }_{\text{eq}}=\frac{\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{\frac{1}{2}+\frac{2}{1}}{\frac{1}{2}+1}=\frac{5}{3}\text{ V}.$$

Equivalent internal resistance of the cells: $$r_{\text{int,eq}}=\frac{r_1{r}_2}{r_1+r_2}=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega .$$

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

Current $$I_2=\frac{{\epsilon }_{\text{eq}}}{\text{total resistance}}=\frac{5/3}{20/3}=\frac{1}{4}\text{ A}.$$

Finally, $$\frac{I_1}{I_2}=\frac{\frac{1}{3}}{\frac{1}{4}}=\frac{4}{3},$$ so $x=4$.