DC response of an RL filter: For the RL network from Problem 50561, if a 15 V DC input is applied and the output is measured across the inductor, what is the steady-state output voltage?

Introduction / Context:
This question checks your understanding of the steady-state DC behavior of inductors in basic RL filters. The output is taken across the inductor, and the input is a constant (DC) voltage source.

Given Data / Assumptions:

• Input is DC: Vin = 15 V (constant).
• RL network as previously described: R in series with L.
• Output node is across the inductor.
• We are interested in steady-state (long after switching transients die out).

Concept / Approach:
At steady-state DC, an ideal inductor behaves like a short circuit (its reactance XL = 2 * pi * f * L is zero at f = 0). Therefore, the DC voltage drop across an ideal inductor is zero after transients have settled.

Step-by-Step Solution:
For DC (f = 0), XL = 0Inductor becomes a short ⇒ V_L(steady) = 0Hence, the steady-state output across the inductor is 0 V

Verification / Alternative check:
Transiently, immediately after application of DC, the inductor initially resists change in current and can have a non-zero voltage. But as time → ∞, current becomes constant and the inductor drop goes to zero.

Why Other Options Are Wrong:

• 15 V: Would require the inductor to be an open at DC, which is false.
• 7.07 V or 10.6 V: These are AC-related magnitudes (rms fractions); not applicable to steady-state DC.

Common Pitfalls:
Confusing instantaneous response with steady-state, or thinking an inductor always blocks or passes voltage like a capacitor. At DC, inductors short; capacitors open.

Final Answer:
0 V