A DC source is connected in series with a 10 kΩ resistor and a capacitor. After a long time (DC steady state), what is the current through the circuit?
AI = V/R — the resistor limits the current just as in a resistive circuit
BI = 0 — in steady state the capacitor acts as an open circuit, blocking DC current
CI = V/Xc — the capacitive reactance limits current at DC
DI depends on the capacitance C and how long the circuit has been running
In DC steady state, the capacitor has fully charged to the source voltage (via KVL: all voltage drops across the open circuit element). Once charged, no current flows — the capacitor acts as a wire break (open circuit). A student who answered A is applying the AC or transient model, where Xc = 1/ωC gives a finite impedance. At DC (ω = 0), Xc → ∞ — truly open. Answer D describes the transient phase that precedes steady state.
Question 2 Multiple Choice
An inductor L is in series with a resistor R connected to a DC source V. In DC steady state, what is the voltage across the inductor?
AV_L = L × I, where I is the steady-state current
BV_L = V (the source voltage) — the inductor stores it as magnetic energy
CV_L = 0 — the inductor acts as a short circuit (ideal wire) in steady state
DV_L = V − IR, where I = V/L
In DC steady state, current is constant (di/dt = 0). The inductor's voltage is V = L(di/dt) = L × 0 = 0. Zero voltage across a component means it behaves like a short circuit — an ideal wire. The steady-state current is then I = V/R (the inductor wire and resistor in series from a voltage source). Option A confuses the inductance formula with resistor behavior. Option B incorrectly applies the transient energy-storage behavior to the steady state.
Question 3 True / False
In DC steady state, a capacitor behaves like an open circuit because it has fully charged and no longer allows current to flow through it.
TTrue
FFalse
Answer: True
True. The capacitor's current equation is i = C(dV/dt). In steady state, nothing is changing — voltages are constant — so dV/dt = 0 and i = 0. No current through a component means it behaves like a wire break (open circuit). The capacitor has charged to whatever voltage the rest of the circuit imposes on it (determined by KVL after replacing it with an open), and it sustains that voltage indefinitely without requiring further current.
Question 4 True / False
In DC steady state, a capacitor has zero voltage across it, because no current flows through it.
TTrue
FFalse
Answer: False
False — this is a critical misconception. The capacitor has zero *current*, not zero voltage. It acts as an open circuit, so the full source voltage (or whatever KVL distributes to it) appears across it. In a simple series RC circuit in steady state, the capacitor charges up to the source voltage: all of V appears across the open-circuit capacitor, not across the resistor (since no current means no resistive voltage drop). Confusing 'open circuit' with 'zero voltage' leads to incorrect operating point calculations.
Question 5 Short Answer
Explain physically why a capacitor acts as an open circuit in DC steady state. Use the relationship i = C(dV/dt) in your answer.
Think about your answer, then reveal below.
Model answer: A capacitor's current is i = C(dV/dt) — it only allows current to flow when the voltage across it is changing. In DC steady state, all voltages and currents have reached constant values (by definition). With dV/dt = 0, the current through the capacitor is exactly zero. A component through which no current flows is — by definition — an open circuit. Physically, the capacitor has charged to the voltage the circuit imposes on it; once charged to that voltage, there is no net electric field driving further charge transfer, so current stops.
The contrast with an inductor is instructive: an inductor's voltage is V = L(di/dt), which is zero when current is constant — making it a short circuit. Capacitor: blocks DC (open circuit). Inductor: passes DC (short circuit). Both rules follow directly from the defining equations of each element, applied at steady state where all time derivatives are zero.