Question:
In the circuit given below, the switch \(S\) was kept open for a sufficiently long time and is closed at time \(t = 0\). The time constant (in seconds) of the circuit for \(t>0\) is \(\_\_\_\_\).
In the circuit given below, the switch \(S\) was kept open for a sufficiently long time and is closed at time \(t = 0\). The time constant (in seconds) of the circuit for \(t>0\) is \(\_\_\_\_\).
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In RL circuits, always find the equivalent resistance first, considering both series and parallel combinations, to compute the time constant accurately.
Updated On: Feb 4, 2026
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Solution and Explanation
Step 1: Calculate the equivalent resistance.
When the switch \(S\) is closed, the two \(4 \, \Omega\) resistors are connected in parallel. The equivalent resistance is: \[ R_{{eq}} = \frac{4 \times 4}{4 + 4} = 2 \, \Omega. \] This equivalent resistance is in series with the \(2 \, \Omega\) resistor, so the total resistance is: \[ R_{{total}} = 2 + 2 = 4 \, \Omega. \] Step 2: Determine the time constant.
The time constant for an RL circuit is given by: \[ \tau = \frac{L}{R_{{total}}}. \] Substituting \(L = 3 \, {H}\) and \(R_{{total}} = 4 \, \Omega\): \[ \tau = \frac{3}{4} = 0.75 \, {seconds}. \] Final Answer: \[ \boxed{{0.75}} \]
When the switch \(S\) is closed, the two \(4 \, \Omega\) resistors are connected in parallel. The equivalent resistance is: \[ R_{{eq}} = \frac{4 \times 4}{4 + 4} = 2 \, \Omega. \] This equivalent resistance is in series with the \(2 \, \Omega\) resistor, so the total resistance is: \[ R_{{total}} = 2 + 2 = 4 \, \Omega. \] Step 2: Determine the time constant.
The time constant for an RL circuit is given by: \[ \tau = \frac{L}{R_{{total}}}. \] Substituting \(L = 3 \, {H}\) and \(R_{{total}} = 4 \, \Omega\): \[ \tau = \frac{3}{4} = 0.75 \, {seconds}. \] Final Answer: \[ \boxed{{0.75}} \]
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