The net current flowing in the given circuit is __________ A.

Solution

Physics sir · 4 steps

In steady DC the capacitor is an open switch, so everything to its right drops out of the circuit.

1. Capacitor blocks DC. Once the 1 μF capacitor is charged, no steady current passes through it. The 2.5 Ω in series with it carries no current, and the 5 Ω, 8 Ω, 1 Ω and 4 Ω on the far side have no source, so they carry none either.
2. Use the bottom wire. The whole bottom line is one node, joined to the negative terminal. The 3 Ω and the 6 Ω both run from the node after the 2 Ω down to this line, so they are in parallel: (3 × 6)/(3+6) = 2 Ω.
3. Two equal branches. That 2 Ω is in series with the top 2 Ω, giving 4 Ω. The diagonal 4 Ω goes straight from the positive terminal to the bottom line, so it sits in parallel with this 4 Ω branch.

   R_(eq) = (4 × 4)/(4+4) = 2 Ω, I = V/R_(eq) = (2 V)/(2 Ω)

4. So. The battery drives I = 1 A. That is 0.5 A through each 4 Ω path. The net current is 1 A.

The common slip: 0.5. Missing that the diagonal 4 Ω sits directly across the battery leaves only one 4 Ω path, which gives 0.5 A instead of 1.

The idea to keep. In a DC circuit at steady state, cut every capacitor out. Any part that is then left without a source carries no current, so you can delete it before you combine resistors.

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