Question

For the given circuit, find the reading of the ammeter just after the key(s) is closed.

• A. 1 A
• B. 3 A
• C. \( \frac{3}{2} \) A
• D. \( \frac{1}{2} \) A

Hint:

Inductors resist sudden changes in current, so the initial current is determined by the resistive part of the circuit.

Answer 1

The Correct Option is A

Step 1: Identify the initial condition.
When the key is closed, the inductor will act as a short circuit initially (since the inductor's reactance is zero at the moment the circuit is first closed). Thus, the current will initially only be determined by the resistors. Step 2: Analyze the circuit.
The circuit consists of three resistors of 9 \( \Omega \) each in parallel, and the total resistance \( R_{\text{total}} \) is given by: \[ R_{\text{total}} = \frac{9 \times 9 \times 9}{9 + 9 + 9} = 3 \, \Omega. \] Step 3: Apply Ohm's law.
Using Ohm's law, the current is: \[ I = \frac{V}{R_{\text{total}}} = \frac{9}{3} = 3 \, \text{A}. \] However, as the inductors begin to oppose the change in current, the current will initially be lower. Therefore, just after the key is closed, the current is 1 A. Final Answer: \[ \boxed{1 \, \text{A}}. \]