Question:medium

The voltage across the inductor for \( t > 0 \) in the given circuit is:

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For an RL circuit, the voltage across the inductor decays exponentially with time, following the equation \( v_L(t) = V_0 e^{-\gamma t} \), where \( \gamma = \frac{R}{L} \).
Updated On: Feb 18, 2026
  • \( v = 25e^{\gamma t} \, \text{V} \)
  • \( v = 25e^{-\gamma t} \, \text{V} \)
  • \( v = -25e^{\gamma t} \, \text{V} \)
  • \( v = -25e^{-\gamma t} \, \text{V} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Inductor Behavior.
In an RL circuit, the inductor voltage for \( t>0 \) is defined by its natural response: \[v_L(t) = V_0 e^{-\gamma t}\] where \( \gamma = \frac{R}{L} \) represents the time constant.
Step 2: Substitute Values.
Given \( V_0 = 25 \, \text{V} \), the inductor voltage for \( t>0 \) is: \[v = 25e^{-\gamma t} \, \text{V}\]
Final Answer: \[\boxed{(2) \, v = 25e^{-\gamma t} \, \text{V}}\]
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