Question

The circuit shown in the figure with the switch \( S \) open is in steady state. After the switch \( S \) is closed, the time constant of the circuit in seconds is:
\includegraphics[width=0.5\linewidth]{14.png}

• A. \(1.25 \)
• B. \(0 \)
• C. \(1 \)
• D. \(1.5 \)

Hint:

For time constants in \( R-L \) networks: - Combine resistances using series/parallel rules. - Combine inductances similarly.

Answer 1

The Correct Option is A

Step 1: Analyzing the circuit after closing the switch. When the switch \( S \) is closed, the circuit becomes an \( R-L \) network. The time constant (\( \tau \)) of an \( R-L \) circuit is given by: \[ \tau = \frac{L_{\text{eq}}}{R_{\text{eq}}}. \] Step 2: Calculating \( R_{\text{eq}} \) (Thevenin resistance). The resistances in the circuit are combined as: \[ R_{\text{eq}} = 1 \, \Omega + 1 \, \Omega = 2 \, \Omega. \] Step 3: Calculating \( L_{\text{eq}} \) (equivalent inductance). The inductors in series are added directly: \[ L_{\text{eq}} = 1+1+\frac{1}{2} = \frac{5}{2} \] Step 4: Determining the time constant (\( \tau \)). \[ \tau = \frac{L_{\text{eq}}}{R_{\text{eq}}} = \frac{5}{2*2} = \frac{5}{4}=1.25 \, \text{seconds}. \]
Hence, the correct option is (1) \(1.25 \).