Question:
The time constant of L-R circuit is
The time constant of L-R circuit is
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Remember: in an L-R circuit, \(\tau = L/R\) (analogous to \(\tau = RC\) in a capacitive circuit). The inductor opposes change in current.
Updated On: Apr 20, 2026
- \(LR\)
- \(\dfrac{L}{R}\)
- \(\dfrac{R}{L}\)
- \(\dfrac{1}{LR}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Time constant \(\tau\) of an L-R circuit: time for current to reach about 63% of final value.
Step 2: Detailed Explanation:
Current growth equation: \(I = I_0(1 - e^{-Rt/L})\). For exponent to be dimensionless: \(\frac{Rt}{L} \Rightarrow \tau = \frac{L}{R}\). Hence, \(\tau\) has dimensions of time \([T]\).
Step 3: Final Answer:
\[ \boxed{\tau = \frac{L}{R}} \]
Time constant \(\tau\) of an L-R circuit: time for current to reach about 63% of final value.
Step 2: Detailed Explanation:
Current growth equation: \(I = I_0(1 - e^{-Rt/L})\). For exponent to be dimensionless: \(\frac{Rt}{L} \Rightarrow \tau = \frac{L}{R}\). Hence, \(\tau\) has dimensions of time \([T]\).
Step 3: Final Answer:
\[ \boxed{\tau = \frac{L}{R}} \]
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