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Question
JEE Main
Physics

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

The correct answer is
2

To determine the reading of the ammeter in steady state, we need to analyze the circuit shown in the image. Here are the steps involved:

  1. In the steady state, a capacitor behaves like an open circuit for DC. Therefore, the 10 μF capacitor can be replaced by an open circuit.
  2. After replacing the capacitor with an open circuit, observe the modified circuit. The branches containing the 4 Ω resistor and the capacitor are open, meaning no current flows through this path.
  3. Now, the circuit simplifies to a combination of resistors in series and parallel.
  4. The 8 Ω resistor and the 8 Ω resistor in parallel with the ammeter form a series circuit with the voltage source.
  5. The equivalent resistance of two parallel 8 Ω resistors is: \(R_{\text{eq}} = \frac{8 \times 8}{8 + 8} = 4 \, \Omega\)
  6. The total resistance in the circuit is a combination of 1 Ω, 8 Ω, and the equivalent resistance of 4 Ω, resulting in: \(R_{\text{total}} = 1 + 8 + 4 = 13 \, \Omega\)
  7. Using Ohm's Law, the total current supplied by the battery is: \(I = \frac{V}{R_{\text{total}}} = \frac{10}{13} \, \text{A}\)
  8. This total current is divided equally between the two parallel branches, each having an 8 Ω resistor. Therefore, the current through the ammeter is: \(I_{\text{ammeter}} = 2 \, \text{A}\)

Hence, the reading of the ammeter in steady state is \(2\) A.

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