A series RL circuit has reached DC steady state with constant current flowing. What is the voltage across the inductor at this moment?
AEqual to the source voltage V — the inductor carries all the voltage at steady state
BEqual to IR — the same as the resistor voltage, since they share current
CZero — because current is constant, di/dt = 0, and v_L = L(di/dt)
DEqual to L/R — the time constant expressed as a voltage
The inductor's defining equation is v_L = L(di/dt). At DC steady state, current is not changing — di/dt = 0 — so the voltage across the inductor is zero regardless of its inductance. The inductor acts as a short circuit (a wire) at DC steady state. All the source voltage appears across the resistor, giving steady-state current I = V/R by Ohm's law. A common mistake is to think the inductor 'holds' voltage permanently — it does not.
Question 2 Multiple Choice
An RL circuit has L = 2 H and R = 4 Ω. A 12 V source is suddenly switched in. What are the time constant τ and the final steady-state current?
Aτ = 8 s, I_final = 3 A
Bτ = 0.5 s, I_final = 3 A
Cτ = 2 s, I_final = 12 A
Dτ = 0.5 s, I_final = 48 A
The time constant is τ = L/R = 2/4 = 0.5 s. The final steady-state current is I_final = V/R = 12/4 = 3 A (at steady state the inductor is a short circuit, so all voltage is across R). At t = τ = 0.5 s, the current has reached about 63% of 3 A ≈ 1.89 A. Option A has the wrong formula (τ = LR instead of L/R); option C has a wrong final current; option D compounds both errors.
Question 3 True / False
Placing a flyback diode across an inductor (e.g., a motor coil or relay) protects circuit components by giving the inductor current a path to flow and dissipate safely when a switch opens.
TTrue
FFalse
Answer: True
When a switch opens abruptly in a circuit carrying steady current through an inductor, the inductor resists the sudden change: v_L = L(di/dt) becomes very large as di/dt spikes. Without a path for the current, this voltage spike can destroy the switch or other components. A flyback diode provides a low-impedance loop for the inductor current to circulate and decay, limiting the voltage spike to the diode's forward voltage (≈0.7 V). This is standard protection for any inductive load.
Question 4 True / False
When a DC source is suddenly connected to a series RL circuit, the current immediately jumps to its final value V/R and then remains constant.
TTrue
FFalse
Answer: False
The inductor resists any instantaneous change in current — this is its fundamental property. At the instant the switch closes (t = 0), the current is zero (assuming it was zero before). It then grows exponentially: i(t) = (V/R)(1 − e^(−t/τ)), reaching V/R only asymptotically. At t = 5τ it is within 1% of V/R and considered at steady state. An instantaneous jump to V/R would require infinite di/dt and thus infinite voltage across the inductor, which is physically impossible.
Question 5 Short Answer
Why does an inductor produce a large voltage spike when a switch suddenly opens a circuit carrying steady current? Use the inductor's defining equation to explain.
Think about your answer, then reveal below.
Model answer: The inductor's defining equation is v_L = L(di/dt). When a switch opens abruptly, the current through the inductor tries to drop from its steady value I₀ to zero in an extremely short time Δt. This means di/dt = −I₀/Δt, which is a very large negative number. Multiplied by L, this produces a very large voltage spike across the inductor — large enough to arc across the switch or destroy components. The inductor is 'fighting' to maintain the current by generating whatever voltage is necessary.
This is the dual of the capacitor's behavior: a capacitor resists instantaneous voltage changes (v = Q/C), while an inductor resists instantaneous current changes (v = L di/dt). The inductive kick is the practical consequence of this resistance: suddenly removing the current path forces v_L to become enormous. Flyback diodes and snubber circuits in motor drives, relay coils, and solenoids are all designed to manage this effect safely.