A 10 mH inductor carries a steady DC current of 2 A. What is the voltage across the inductor?
A20 mV, calculated as L × I
B0 V, because di/dt = 0 for constant current
C200 V, because the inductor stores energy equal to ½LI²
DIt depends on the frequency of the AC source driving the circuit
The inductor's voltage-current relationship is v = L(di/dt). If current is constant (steady DC), then di/dt = 0 and v = 0. An ideal inductor with constant current looks like a short circuit — a perfect wire with no voltage drop. The stored energy ½LI² exists in the magnetic field, but energy storage doesn't produce a voltage; only *changing* current does. Option A (L × I) confuses the energy formula with the voltage formula.
Question 2 Multiple Choice
As the frequency of an AC signal increases, how do the impedances of an inductor and a capacitor change relative to each other?
The inductor and capacitor are duals with opposite frequency behavior. Z_L = jωL grows with ω — at high frequencies, the inductor strongly opposes current (high impedance). Z_C = 1/(jωC) shrinks with ω — at high frequencies, the capacitor offers little opposition (low impedance). This opposing behavior is what makes LC circuits oscillate: at high frequencies the capacitor is easy but the inductor is hard; at low frequencies the reverse is true. The balance point is the resonant frequency.
Question 3 True / False
An ideal inductor carrying a constant DC current has zero voltage across its terminals.
TTrue
FFalse
Answer: True
Since v = L(di/dt) and a constant DC current has di/dt = 0, the voltage is exactly zero. The inductor acts like a short circuit for DC. This is why ideal inductors are invisible to steady-state DC analysis — they appear as wires. The magnetic energy ½LI² is stored in the field but does not produce a terminal voltage unless the current is changing.
Question 4 True / False
In a DC circuit, a capacitor and an inductor behave the same way — both appear as short circuits in steady state.
TTrue
FFalse
Answer: False
They are opposites in DC steady state — a precise manifestation of their duality. A capacitor blocks DC current: in steady state, current stops flowing through it (i = C·dv/dt = 0 when v is constant), making it an open circuit. An inductor passes DC current freely: once current stabilizes, v = L·di/dt = 0, making it a short circuit. Confusing these two behaviors leads to serious errors in circuit analysis, particularly in power supply and filter design.
Question 5 Short Answer
Explain why you cannot instantaneously change the current through an inductor, and what physical consequence occurs if a circuit attempts to do so.
Think about your answer, then reveal below.
Model answer: The inductor's voltage is v = L(di/dt). Instantaneously changing current requires di/dt → ∞, which would require infinite voltage — physically impossible. If a switch abruptly breaks a circuit carrying inductor current, the inductor briefly sustains the current by generating a large voltage spike (voltage as high as the circuit allows). In practice this can cause arcing across the switch or destroy unprotected components.
This property — current continuity — is the inductor's most important circuit behavior. It is the electromagnetic analog of mechanical inertia: just as you cannot instantaneously stop a moving mass, you cannot instantaneously stop current in an inductor. The stored magnetic energy ½LI² must go somewhere; if the circuit doesn't provide a path, the inductor creates one through a voltage spike. Freewheeling diodes in motor driver circuits exploit this principle to safely absorb the energy when current is switched off.