Q74 / 75 · JEE Main 2026 (05 April Shift 2) · Physics
When an external resistance of $5\ \Omega$ is connected across terminals of a cell, a current of $0.25\text{ A}$ flows through it. When the $5\ \Omega$ resistor is replaced by a $2\ \Omega$ resistor, a current of $0.5\text{ A}$ flows through it. The internal resistance of the cell is _______ $\Omega$.
Correct answer: 1
Solution
Let the EMF of the cell be $E$ and its internal resistance be $r$.
The current $I$ in the circuit is given by $I = \dfrac{E}{R + r}$.
For the first case, $R_1 = 5\ \Omega$ and $I_1 = 0.25\text{ A}$:
$0.25 = \dfrac{E}{5 + r} \Rightarrow E = 0.25(5 + r)$
For the second case, $R_2 = 2\ \Omega$ and $I_2 = 0.5\text{ A}$:
$0.5 = \dfrac{E}{2 + r} \Rightarrow E = 0.5(2 + r)$
Equating the expressions for $E$:
$0.25(5 + r) = 0.5(2 + r)$
$5 + r = 2(2 + r)$
$5 + r = 4 + 2r$
$r = 1\ \Omega$
Answer: $1$
The current $I$ in the circuit is given by $I = \dfrac{E}{R + r}$.
For the first case, $R_1 = 5\ \Omega$ and $I_1 = 0.25\text{ A}$:
$0.25 = \dfrac{E}{5 + r} \Rightarrow E = 0.25(5 + r)$
For the second case, $R_2 = 2\ \Omega$ and $I_2 = 0.5\text{ A}$:
$0.5 = \dfrac{E}{2 + r} \Rightarrow E = 0.5(2 + r)$
Equating the expressions for $E$:
$0.25(5 + r) = 0.5(2 + r)$
$5 + r = 2(2 + r)$
$5 + r = 4 + 2r$
$r = 1\ \Omega$
Answer: $1$
Hence, the correct answer is 1.