The reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ____ A.

Solution

Physics sir · 4 steps

In steady state the capacitor branch carries nothing, so remove it and reduce the resistors that remain.

1. Drop the capacitor. In steady state a capacitor is fully charged and no current flows through it. The diagonal branch (10 μF in series with 8 Ω) carries zero current, so you can remove it.
2. Right-hand path. The two right 8 Ω resistors are in parallel, giving 4 Ω. Add the 4 Ω on top in series and this path is 8 Ω. It returns to the bottom wire through the ammeter.
3. Total current. This 8 Ω path is in parallel with the left 8 Ω resistor, which gives 4 Ω. Add the 1 Ω in series, so the total resistance is 5 Ω.

   I = (10 V)/(1 Ω + 4 Ω) = 2 A

4. So. The 2 A splits equally between two equal 8 Ω paths. The right path, the one through the ammeter, carries 1 A, so the reading is 1 A, option (b).

The idea to keep. In a DC steady state, treat every capacitor as an open switch. Then find which branch the ammeter sits in and work out how the total current divides into it.

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