In the given circuit, the current flowing through the resistance 20 Ω is 0.3 A, while the ammeter reads 0.9 A. The value of R₁ is _________ Ω.

Solution

Physics sir · 4 steps

All three resistors share the same voltage, so the 20 Ω branch gives you that voltage straight away.

1. Spot the parallel. R₁, 20 Ω and 15 Ω all join the same two junctions, so they are in parallel. The ammeter sits after the junction and reads the total current, 0.9 A.
2. Common voltage. Across the 20 Ω branch, V = 0.3 × 20 = 6 V. The same 6 V is across every branch.
3. Other branch currents. Through 15 Ω: 6/15 = 0.4 A. By the junction rule, the rest of the 0.9 A goes through R₁.

   I₁ = 0.9 - 0.3 - 0.4 = 0.2 A

4. So. R₁ = 6/0.2 = 30 Ω, so the answer is 30.

The common slip: 10. Forgetting the 15 Ω branch and giving all of the remaining 0.6 A to R₁ gives 6/0.6 = 10 Ω, not 30.

The idea to keep. In a parallel network, one known branch current gives you the common voltage. The junction rule then hands you whatever current is missing.

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