As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3 Ω. The currents through the resistors R₄ and R₅ are I₄ and I₅ respectively. The values of I₄ and I₅ are :

Solution

Physics sir · 4 steps

Reduce the whole circuit to one resistance, find the main current, then split it between R4 and R5.

1. Combine the parallels. R₁ ∥ R₂ = (2 × 2)/(2 + 2) = 1 Ω. R₄ ∥ R₅ = (20 × 5)/(20 + 5) = 4 Ω.
2. Total resistance. Everything else is in series with these, including the internal resistance: 1 + 2 + 4 + 2 + 3 = 12 Ω. So the main current is I = 24/12 = 2 A.
3. Split the current. In two parallel branches, each branch carries the current in proportion to the other branch's resistance. The smaller resistor gets the larger share.

   I₄ = I × R₅/(R₄ + R₅) = 2 × 5/25 = 2/5 A, I₅ = 2 × 20/25 = 8/5 A

4. So. I₄ = 2/5 A and I₅ = 8/5 A, which is option (c). Check: 2/5 + 8/5 = 2 A, the full main current.

The idea to keep. For a series chain of parallel blocks, collapse each block, include the internal resistance, find the main current, then split it inversely with resistance.

Answer from the NTA final answer key. Solution by PrepWiser.