The voltage drop across 15 Ω resistance in the given figure will be V. [Figure: two branches in parallel between a left node and a right node. Upper branch in series: (4 Ω∥4 Ω), then 2 Ω, then (15 Ω∥10 Ω). Lower branch in series: (8 Ω∥8 Ω), then (12 Ω∥12 Ω). The left and right nodes are joined thro…

Question

The voltage drop across 15 Ω resistance in the given figure will be V. [Figure: two branches in parallel between a left node and a right node. Upper branch in series: (4 Ω∥4 Ω), then 2 Ω, then (15 Ω∥10 Ω). Lower branch in series: (8 Ω∥8 Ω), then (12 Ω∥12 Ω). The left and right nodes are joined through a 12 V cell in series with 1 Ω.]

Accepted Answer

6. Upper branch: 2+2+(15×10)/(25)=2+2+6=10 Ω. Lower branch: 4+6=10 Ω. Branches in parallel: 5 Ω; with the 1 Ω: 6 Ω, so I=(12)/6=2 A. Each branch carries 1 A. Voltage across 15 Ω∥10 Ω =1×6=6 V.

Step-by-step solution