In a DC steady-state circuit, what is the correct equivalent circuit model for a capacitor?
AA short circuit — capacitors pass DC current freely once charged
BAn open circuit — no DC current flows through a capacitor in steady state
CA resistor whose value is 1/(ωC), where ω is the DC frequency
DA voltage source equal to the initial charge on the capacitor
A capacitor's current is I = C · dV/dt. In DC steady state, all voltages are constant, so dV/dt = 0 everywhere. Therefore I = 0: no current flows through the capacitor. A device that passes no current is by definition an open circuit. This substitution rule — replace capacitors with open circuits — is the key tool for DC steady-state analysis. Note that the capacitor can still have a non-zero voltage across it; the open-circuit substitution means no current flows, not that no voltage exists.
Question 2 Multiple Choice
You want to find the voltage across a capacitor in a DC steady-state circuit. A classmate suggests solving the full differential equation for V(t) and then taking the limit as t → ∞. What is the faster correct approach?
AReplace the capacitor with a short circuit and solve the resistor network for current through that branch
BReplace the capacitor with an open circuit and solve the remaining resistor network for the voltage at those terminals
CUse phasor analysis with ω = 0 to find the DC impedance of the capacitor
DApply the voltage divider rule using the capacitor's DC impedance
In DC steady state, a capacitor is simply an open circuit — no current flows through it. To find its voltage, replace it with an open circuit and solve the remaining resistor-source network for the open-circuit terminal voltage. Whatever voltage appears across those open terminals is the capacitor's steady-state voltage. This avoids solving any differential equation. Using a short circuit (option A) is the rule for inductors, not capacitors. Phasors and impedance (option C) are AC steady-state tools; at DC (ω = 0), capacitor impedance 1/jωC → ∞, which confirms the open-circuit result, but the direct substitution approach is far simpler.
Question 3 True / False
In DC steady state, an inductor acts as a short circuit because V = L·dI/dt equals zero when current is not changing.
TTrue
FFalse
Answer: True
True. An inductor's terminal voltage is V = L · dI/dt. In DC steady state, all currents are constant (no time variation), so dI/dt = 0 and therefore V = 0. A circuit element with zero voltage across it is a short circuit — it behaves like a wire. The inductor passes DC current freely with no voltage drop. This is the complement of the capacitor rule: capacitors → open circuits (no current), inductors → short circuits (no voltage). Both follow directly from the condition dV/dt = dI/dt = 0.
Question 4 True / False
A capacitor in DC steady state cannot have a nonzero voltage across it, because capacitors block DC current and therefore can seldom store energy in a DC circuit.
TTrue
FFalse
Answer: False
False. A capacitor in DC steady state can absolutely have a nonzero voltage across it — in fact, finding that voltage is often the goal of DC steady-state analysis. What is zero in steady state is the *current* through the capacitor (I = C dV/dt = 0). The capacitor holds whatever voltage was established during the transient charging phase. 'Blocks DC current' means zero steady-state current, not zero steady-state voltage. The stored energy ½CV² is nonzero whenever V ≠ 0.
Question 5 Short Answer
Explain why a capacitor behaves as an open circuit in DC steady state, and state the physical condition that must hold for this to be true.
Think about your answer, then reveal below.
Model answer: A capacitor's current obeys I = C · dV/dt. In DC steady state, the circuit has reached a time-invariant condition where all voltages and currents are constant — therefore dV/dt = 0 everywhere. Substituting, I = C · 0 = 0: no current flows through the capacitor. By definition, an element through which no current flows is an open circuit. The physical condition that must hold is that the circuit has fully settled after all transients have decayed — no part of the circuit is still changing. During the transient (e.g., right after a switch closes), dV/dt ≠ 0 and the capacitor does carry current.
This is why DC steady-state analysis is sometimes called 'long-time analysis' — it only applies after t → ∞, once all exponential transients have died out. The substitution rules are not approximations; they are exact consequences of dV/dt = dI/dt = 0.