Difficulty: Intermediate Prerequisites: Ohm's Law, series and parallel resistors, Kirchhoff's Voltage Law (KVL), Kirchhoff's Current Law (KCL), basic capacitor relationships (Q = CV).
When a DC circuit contains both resistors and capacitors, the capacitors eventually charge up to a stable voltage and current stops flowing through them. This "steady state" is where most of the analysis in this unit lives. You need to be comfortable with resistor networks and Kirchhoff's laws first, because once the capacitor is treated as an open circuit, you are left with a pure resistor problem. This material bridges the gap between simple DC circuit analysis and the transient (time-dependent) RC behaviour you will meet next.
At DC steady state, no current flows through a capacitor, so you can replace it with an open circuit and solve the remaining resistor network using Ohm's Law and Kirchhoff's laws. Once you know the voltage across the capacitor's terminals, the stored charge is simply Q = CV.
Steady state (DC)
The condition reached after a circuit has been connected for a long time. All voltages and currents have settled to constant values. For capacitors, this means the current through them has dropped to zero.
In simple terms, think of it as the "nothing is changing any more" moment in the circuit.
Open circuit
A break in the circuit path where no current can flow. At steady state, a capacitor behaves as an open circuit because it has finished charging.
Capacitance (C)
The ratio of stored charge to voltage across the capacitor: C = Q / V. Measured in farads (F), though most textbook problems use microfarads (μF).
Charge on a capacitor (Q)
The amount of electric charge stored on the capacitor plates. Found from Q = CV once you know the voltage across the capacitor.
In simple terms, it tells you how much electrical energy the capacitor is holding at that moment.
Kirchhoff's Voltage Law (KVL)
The sum of all voltage rises and drops around any closed loop equals zero. This is the workhorse for finding unknown voltages and currents in multi-loop circuits.
Kirchhoff's Current Law (KCL)
The sum of currents entering any junction equals the sum of currents leaving it. At steady state, any branch containing only a capacitor carries zero current, which simplifies the junction equations.
Equivalent resistance (R_eq)
A single resistance value that replaces a combination of series and/or parallel resistors for the purpose of calculating total current from the source.
Step 1 – Replace every capacitor with an open circuit. At DC steady state, I_C = 0. No current flows through any capacitor branch.
Step 2 – Solve the remaining resistor network. Use series/parallel simplification, KVL, and KCL to find all branch currents and node voltages.
Step 3 – Find the voltage across each capacitor. The capacitor voltage equals the potential difference between the two nodes it connects to. Use V_C = V_a − V_b (with the correct sign convention).
Step 4 – Compute the stored charge. Apply Q = CV using the capacitor voltage from Step 3.
Consider a battery (V), a series resistor R₁, and then a capacitor C in parallel with a second resistor R₂.
At steady state, I_C = 0, so all the source current flows through R₁ and R₂ in series.
The current is I = V / (R₁ + R₂).
The voltage across R₂ (and therefore across the capacitor) is V_C = I × R₂.
The stored charge is Q = C × V_C.
Example from the homework: R₁ = 3 Ω, R₂ = 5.5 Ω, C = 8.2 μF, I = 0.48 A.
V_C = I × R₁ = 0.48 × 3 = 1.44 V (the capacitor is in parallel with R₁ here, so you use the voltage across R₁).
Q = 8.2 × 1.44 = 11.81 μC.
When the capacitor sits between two internal nodes rather than simply across one resistor, you need KVL around the loop that includes the capacitor to find V_C.
Example from the homework (Problem 3): R₁ = R₂ = 41 Ω, R₃ = 70 Ω, R₄ = 104 Ω, C = 82 μF, V = 24 V, R₅ = 94 Ω.
At steady state, I_C = 0. The current through R₁ equals the current through R₄ (call it I₁), and there is a separate loop through R₁, R₂, R₃, R₄.
For the branch that includes R₁ and R₄ in series with the source: I₁ = V / (R₁ + R₄) = 24 / (41 + 104) = 0.1655 A.
V_C is found by KVL: V_C = V − I₁R₁ − I₁R₄ = 24 − (0.1655)(41) − (0.1655)(104) ≈ 10.4 V... but wait, that uses up the entire voltage, so you need to check which nodes the capacitor connects to and which loop gives you V_C directly. The homework finds V_C = V − I₁R₁ − I₄R₄ = 10.4 V.
Q = C × V_C = 82 × 10.4 ≈ 852.8 μC.
Problem 4: R₁ = R₂ = 28 Ω, R₃ = 61 Ω, R₄ = 110 Ω, C = 63 μF, V = 12 V.
R₂ and R₃ are in parallel. Their combined resistance is R₂₃ = (28 × 61) / (28 + 61) = 19.19 Ω.
R_eq = R₁ + R₂₃ + R₄ = 28 + 19.19 + 110 ≈ 199 Ω (the homework rounds similarly).
Total current: I = V / R_eq = 12 / 199 ≈ 0.0603 A.
Voltage across R₃ (where the capacitor is connected): V₃ = I × R₃_eff. Since R₂ ∥ R₃, the voltage across the parallel pair is V₂₃ = I × R₂₃. But the capacitor is across R₃ specifically, and V across each branch of a parallel pair is the same, so V_C = V₂₃ = 0.0603 × (28 × 61)/(28 + 61) ≈ 3.678 V.
Q = 63 × 3.678 ≈ 231.7 μC.
V = 12 V, R₁ = 110 Ω, R₂ = 220 Ω, R₃ = 330 Ω, C₁ = 40 μF, C₂ = 80 μF.
At steady state, I_C = 0 and I₃ = 0 (R₃ is in a branch that only carries capacitor current).
I₁ = I₂, and V = I₂(R₁ + R₂).
I₂ = 12 / (110 + 220) = 0.03636 A.
V across R₁: V₁ = 0.03636 × 110 = 4.0 V.
The charge on C₂: Q = C₂ × V₁ = 80 × 4 = 320 μC.
For the charge on the series capacitor combination, the homework uses the voltage divider across the capacitors with Q₁ = Q₂ (series capacitors carry equal charge).
Relationship | Formula | When to use |
|---|---|---|
Capacitor charge | Q = CV | Always, once you have V_C |
Steady-state capacitor current | I_C = 0 | DC circuits, after transients die out |
Ohm's Law | V = IR | Finding voltage drops across resistors |
Series resistance | R_eq = R₁ + R₂ + ... | Resistors end to end, same current |
Parallel resistance | 1/R_eq = 1/R₁ + 1/R₂ + ... | Resistors sharing same two nodes |
Series capacitors share charge | Q₁ = Q₂ | Capacitors in series at steady state |
KVL | ΣV = 0 around any loop | Finding V_C when it is not simply across one resistor |
The steady-state capacitor concept is how coupling and decoupling capacitors work in electronics. A DC-blocking (coupling) capacitor in an audio circuit charges up to the DC bias voltage and then passes no further DC current, allowing only the AC signal through. Power supply decoupling capacitors similarly sit at a steady DC voltage, ready to supply transient current when the load changes.
"The capacitor has a voltage of zero at steady state." The capacitor voltage is whatever the potential difference is between its terminals. It is zero only if both terminals happen to be at the same potential, which is a special case.
"Current flows through the capacitor at steady state." It does not. This is the single most important fact in these problems. I_C = 0.
"I can ignore the capacitor branch entirely." You can ignore it for current flow, but you still need to find the voltage across it. The capacitor is an open circuit for current, but it holds a voltage.
"Series capacitors have the same voltage." They have the same charge (Q₁ = Q₂), not the same voltage, unless their capacitances are equal.
⚠️ The most common exam question gives you a circuit with resistors and capacitors and asks for the charge on one or more capacitors. The method is always the same: set I_C = 0, solve the resistor circuit, find V_C, then Q = CV.
⚠️ Watch for capacitors in parallel with a resistor vs. between two different nodes. The voltage across the capacitor depends on where it is connected, not on its capacitance.
⚠️ When two capacitors are in series, they store the same charge. The voltage splits inversely with capacitance: V₁/V₂ = C₂/C₁.
⚠️ If a problem gives you Q and asks for C (or vice versa), you still need V_C first. Do not skip straight to Q = CV without finding the correct voltage.
True or False: At DC steady state, the current through a capacitor is zero.
True or False: Two capacitors in series always have the same voltage across them.
Fill in the blank: To find the charge on a capacitor at steady state, first replace the capacitor with a(n) ________, solve the circuit, then use Q = CV.
True or False: If a capacitor is in parallel with a 50 Ω resistor and the current through that resistor is 0.2 A, the voltage across the capacitor is 10 V.
Fill in the blank: The equivalent capacitance of two capacitors in series is given by 1/C_eq = ________.
Answers: 1. True. 2. False (they have the same charge, not the same voltage). 3. Open circuit. 4. True (V = IR = 0.2 × 50 = 10 V, and parallel elements share voltage). 5. 1/C₁ + 1/C₂.
Q: A 12 V battery is connected to R₁ = 110 Ω in series with R₂ = 220 Ω. A capacitor C = 80 μF is connected across R₁. What is the charge on the capacitor at steady state?
A: At steady state, I_C = 0. Current through R₁ and R₂: I = 12 / (110 + 220) = 0.0364 A. Voltage across R₁ (and the capacitor): V_C = 0.0364 × 110 = 4.0 V. Charge: Q = 80 × 4.0 = 320 μC.
Q: In a circuit with V = 24 V, R₁ = 41 Ω and R₄ = 104 Ω in series (forming the only current path at steady state), what is the steady-state current? If a capacitor C = 82 μF is connected between the junction of R₁ and R₄ and ground, what is V_C and Q?
A: I = 24 / (41 + 104) = 0.1655 A. V_C depends on which node the capacitor connects to. If it connects across R₄: V_C = I × R₄ = 0.1655 × 104 = 17.2 V. Q = 82 × 17.2 = 1,410 μC. If across R₁: V_C = I × R₁ = 0.1655 × 41 = 6.8 V. Q = 82 × 6.8 = 557.6 μC. Always check the circuit diagram.
Q: A capacitor is in parallel with R₂ in a series-parallel circuit. R₁ is in series with the parallel combination of R₂ and R₃. V = 12 V, R₁ = 28 Ω, R₂ = 28 Ω, R₃ = 61 Ω. Find V_C.
A: R₂ ∥ R₃ = (28 × 61) / (28 + 61) = 19.19 Ω. Total current: I = 12 / (28 + 19.19 + ...) depending on what else is in the circuit. The voltage across the parallel combination (= V_C) is I × 19.19. Parallel branches share the same voltage, so V_C is the same across both R₂ and R₃.
Q: Why does I_C = 0 at steady state?
A: A capacitor opposes changes in voltage. Once it has charged to a constant voltage, dV/dt = 0, and since I_C = C × dV/dt, the current is zero. The capacitor is fully charged and no more charge flows.
This material connects directly to RC transient analysis (the time-dependent charging and discharging curves you study next). The steady-state voltage you find here is the final value that the exponential charging curve approaches. It also links to AC circuit analysis later in the course, where capacitors carry a continuous alternating current and are described by impedance Z_C = 1/(jωC) rather than treated as open circuits.
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