Time-constant control in RL networks: Which action will decrease (reduce) the time constant τ of a simple series RL circuit, where τ = L / R?

Introduction / Context:The time constant τ characterizes how fast current and voltage waveforms rise or decay in first-order circuits. For the series RL network, τ = L / R. Designers often need to speed up or slow down responses—for instance, reducing relay kick times or shaping edges in pulse networks—by adjusting L or R appropriately.

Given Data / Assumptions:

• Simple single-loop series RL circuit.
• Time constant τ = L / R (L in henries, R in ohms).
• Objective: reduce τ (achieve a faster response).

Concept / Approach:Because τ = L / R, there are two direct levers: decrease L or increase R. Increasing R reduces τ (faster decay and rise to steady state), while increasing L increases τ (slower response). Changing input amplitude does not affect τ, and swapping component positions in a single loop has no effect on τ.

Step-by-Step Solution:Formula: τ = L / R.To reduce τ, either lower L or raise R.Option selected: add series resistance → R_total increases → τ decreases.Therefore, the circuit responds faster (shorter transient time).

Verification / Alternative check:Simulate with R = 100 Ω, L = 10 mH → τ = 0.01 / 100 = 0.0001 s. Add 100 Ω in series → R = 200 Ω → τ = 0.01 / 200 = 0.00005 s, exactly half, confirming the effect.

Why Other Options Are Wrong:Adding an inductor in series: increases L, increasing τ (slower response).Decreasing input amplitude: scales voltages/currents but does not change τ.Exchanging component positions: in a series loop, ordering does not change τ.

Common Pitfalls:Trying to reduce τ by putting a resistor in parallel with the series resistor; that actually lowers equivalent R and increases τ. Always check whether the modification raises or lowers the series resistance seen by the inductor.

Final Answer:Adding an extra resistor in series to increase the total resistance