The voltage drop across 15Ω resistance in the given figure will be ______________ V.

Solution

Physics sir · 4 steps

Shrink each parallel pair first. The two branches come out equal, so the current splits evenly.

1. Top branch. The two 4 Ω in parallel give 2 Ω. The 15 Ω and 10 Ω in parallel give (15 × 10)/25 = 6 Ω. Adding the middle 2 Ω, the top branch is 2 + 2 + 6 = 10 Ω.
2. Bottom branch. The two 8 Ω give 4 Ω and the two 12 Ω give 6 Ω. In series that is 10 Ω.
3. Total current. Two 10 Ω branches in parallel give 5 Ω. With the 1 Ω in series, the circuit is 6 Ω, so I = 12/6 = 2 A. The equal branches take 1 A each.

   V_(15 Ω) = Iₜₒₚ × R_(15∥ 10) = 1 A × 6 Ω = 6 V

4. So. The 15 Ω sits in parallel with the 10 Ω, so it has the same 6 V across it. The answer is 6.

The common slip: 7.2. Leaving out the 1 Ω in the battery line makes the current 2.4 A, so each branch carries 1.2 A and the drop comes out as 7.2 V.

The idea to keep. Collapse parallel pairs, add series parts, find the total current, then walk back. Resistors in parallel share one voltage, so find the drop across the pair.

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