When a 5 Ω external resistor is connected across a cell, the current is 0.25 A. When the resistor is replaced by a 2 Ω resistor, the current becomes 0.50 A. If the internal resistance of the cell is N ohms, find N.
Physics · Current electricity · circuit theory
JEE Mains 2026 5 April Evening Shift

Question

When a 5 Ω external resistor is connected across a cell, the current is 0.25 A. When the resistor is replaced by a 2 Ω resistor, the current becomes 0.50 A. If the internal resistance of the cell is N ohms, find N.

Physics · Current electricity · circuit theory

JEE Mains 2026 5 April Evening Shift

G1 Physics · Accepted Answer

Let the cell emf be E and internal resistance be r. With the 5 Ω resistor: 0.25 = E ÷ (5+r). Therefore, E = 0.25(5+r). (1) With the 2 Ω resistor: 0.50 = E ÷ (2+r). Therefore, E = 0.50(2+r). (2) Equate (1) and (2): 0.25(5+r) = 0.50(2+r). Thus, 1.25+0.25r = 1+0.50r. Therefore, 0.25 = 0.25r, r = 1 Ω. Hence N = 1.