In the figure shown below, a resistance of 150.4 Ω is connected in series to an ammeter A of resistance 240 Ω. A shunt resistance of 10 Ω is connected in parallel with the ammeter. The reading of the ammeter is___________mA .

Solution

Physics sir · 4 steps

Find the total current first, then split it: the ammeter only reads its own share, not the shunt's.

1. Combine the parallel pair. The ammeter (240 Ω) and the shunt (10 Ω) are in parallel. Their combined resistance is (240 × 10)/(240 + 10) = 2400/250 = 9.6 Ω.
2. Total current. This 9.6 Ω is in series with 150.4 Ω, giving 160 Ω in all. The cell drives I = 20/160 = 0.125 A through the circuit.

   I_A = I × Rₛₕᵤₙₜ/(R_A + Rₛₕᵤₙₜ) = 0.125 × 10/250

3. Split the current. In parallel, current divides in inverse ratio of resistance. The ammeter's resistance is 24 times the shunt's, so it takes only 10/250 = 1/25 of the total.
4. So. I_A = 0.125/25 = 0.005 A, so the ammeter reads 5 mA.

The common slip: 125. Taking the total circuit current as the reading gives 125 mA. Most of that current flows through the 10 Ω shunt, not through the ammeter.

The idea to keep. A meter reads only the current through itself. With a shunt across it, find the total current, then use the current divider: each branch gets the other branch's resistance over the sum.

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