Physics · Current Electricity

JEE Main 2025 — 3 April, Evening Shift — Question 65

Two cells of emfs 1 V and 2 V and internal resistances $2 \Omega$ and $1 \Omega$ , respectively, are connected in series with an external resistance of $6 \Omega$ . The total current in the circuit is $\Lambda_{1}$ . 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 $\left(\frac{l_{1}}{l_{2}}\right)$ is $\frac{x}{3}$ . The value of $x$ is

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

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

{{I}_{2}}

=\frac{\frac{{{\varepsilon }_{1}}}{{{r}_{1}}}+\frac{{{\varepsilon }_{1}}}{{{r}_{2}}}}{\frac{\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}}{\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}+R}}=\frac{\frac{\frac{1}{2}+\frac{2}{1}}{\frac{1}{2}+\frac{1}{1}}}{\frac{1\times 2}{3}+6}=\frac{5\times 2\times 3}{2\times 3\times 20}

=\frac{1}{4}

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

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge

Answer: 4

Step-by-step solution

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