Difficulty: Easy
Correct Answer: Correct
Explanation:
Introduction / Context:Time constant governs how quickly currents and voltages rise or decay during switching events in first-order circuits. For RL networks, the time constant directly affects relay delays, inrush current, and filter settling. This question asks whether resistance influences that time constant.
Given Data / Assumptions:
Concept / Approach:The RL time constant is τ = L / R (seconds). Larger resistance speeds up decay (smaller τ), while larger inductance slows it down (larger τ). This emerges from the differential equation: L * di/dt + R * i = V_step, whose homogeneous solution is i(t) = I_final * (1 − exp(−t/τ)) for a rise or i(t) = I_initial * exp(−t/τ) for a decay.
Step-by-Step Solution:
1) Write the KVL for the series RL under a step: L * di/dt + R * i = V.2) Identify the coefficient of i: R / L sets the exponential rate.3) Conclude τ = L / R and note its units: henry/ohm = second.4) Interpret physically: more resistance dissipates energy faster, so the inductor current reaches the new steady state sooner.Verification / Alternative check:Check limits: as R → 0, τ → ∞ (current changes very slowly); as R → ∞, τ → 0 (current cannot build, response is fast).
Why Other Options Are Wrong:Incorrect or “depends only on inductance”: both ignore the R term that sets the decay rate. Depends only on source voltage: voltage scales final current, not the exponential time constant.
Common Pitfalls:Confusing RL (τ = L/R) with RC (τ = R*C); forgetting that any series resistance adds to R, including winding and source resistances.
Final Answer:Correct
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