JEE Main202324 Jan 2023Morning ShiftPhysicsCurrent ElectricityActual
As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3 Ω . The currents through the resistors R 4 and R 5 are I 4 and I 5 respectively. The values of I 4 and I 5 are:
Options
- AI 4 = 8 5   A and I 5 = 2 5   A
- BI 4 = 24 5   A and I 5 = 6 5   A
- CI 4 = 6 5   A and I 5 = 24 5   A
- DI 4 = 2 5   A and I 5 = 8 5   A
Correct answer
D. I 4 = 2 5   A and I 5 = 8 5   A
Step-by-step solution
Equivalent resistance of circuit R eq = 3 + 2 × 2 2 + 2 + 2 + 20 × 5 20 + 5 + 2 = 12   Ω Current through battery i = 24 12 = 2   A Now, I 4 = R 5 R 4 + R 5 × 2 = 5 20 + 5 × 2 = 2 5   A and I 5 = i - I 4 = 2 - 2 5 = 8 5   A