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

To find the reading of the ammeter just after the key is closed, we need to analyze the circuit given in the diagram. Here, we have a combination of resistors, an inductor, and a voltage source.

Let's break down the circuit:

1. Just after the key is closed, the inductor behaves like an open circuit. Inductors oppose changes in current, and initially, they do not allow current to pass through when the current is suddenly applied (at t = 0).
2. The circuit simplifies to only the resistors since the inductors are an open circuit initially.
3. We have three 9Ω resistors in parallel. The equivalent resistance \( R_{\text{eq}} \) of resistors in parallel is calculated as:
   \[ \frac{1}{R_{\text{eq}}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \]
   Hence, \( R_{\text{eq}} = 3 \, \Omega \).
4. The voltage source is 9V. Using Ohm's Law \( V = IR \), we can find the current \( I \) through the equivalent resistance:
   \[ I = \frac{V}{R_{\text{eq}}} = \frac{9}{3} = 3 \, \text{A} \]
5. Since the inductors are initially open circuits (no current flows through them), the ammeter reading is the current through the resistors. Therefore, the ammeter reads 3 A.

Hence, the correct option is 3 A.