On this page
A capacitor in a DC circuit with resistors looks as if it should make the problem harder. Once the circuit has settled, it does the opposite: no current flows in the capacitor's branch, so you can take the capacitor out, solve a plain resistor circuit, and put it back only to read off its voltage. This post explains why, works three circuits with numbers, and lists the mistakes that cost marks.
Which exams test this?
The JEE (Main) 2026 syllabus lists capacitors in series and parallel and the energy stored in a capacitor (Unit 11), and Kirchhoff's laws (Unit 12). The JEE (Advanced) 2026 brochure goes further and lists "RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources". The CBSE Physics curriculum for 2026-27 covers capacitors and Kirchhoff's rules in Class XII. Check the syllabus for your own exam year.
What does steady state mean for a capacitor?
When a switch closes, charge flows onto the plates and the capacitor's voltage rises. As it rises, the current into it falls. Long afterwards, in practice after several time constants, the capacitor's voltage equals the potential difference between the two points it is connected to, and its current is zero. The NCERT chapter Electrostatic Potential and Capacitance defines capacitance as C = Q/V, which is the only formula you need once you know V.
Two rules follow:
At steady state a capacitor is a gap for DC. No current flows through it, or through anything in series with it.
Its voltage is the potential difference between its two nodes. Then Q = CV.
The opposite holds just after the switch closes. An uncharged capacitor has zero voltage, so for that instant it behaves like a plain wire. Our post on RC circuits and the time constant covers what happens in between.
Worked example 1: a capacitor across one resistor
A battery of EMF 18 V and internal resistance 1 Ω drives R₁ = 3 Ω and R₂ = 5 Ω in series. A 4 µF capacitor is connected across R₂. Find its charge at steady state.
Remove the capacitor. Total resistance = 1 + 3 + 5 = 9 Ω, so I = 18 / 9 = 2 A.
Voltage across R₂ = 2 × 5 = 10 V. This is the capacitor's voltage.
Q = CV = 4 µF × 10 V = 40 µC. Energy stored U = ½CV² = ½ × 4 × 10⁻⁶ × 10² = 2.0 × 10⁻⁴ J, or 200 µJ.
If you had used the full 18 V you would get 72 µC. The capacitor only sees the voltage across the resistor it sits beside.
Worked example 2: a capacitor bridging two branches
Two branches are connected in parallel across an ideal 12 V battery. Branch A has 2 Ω on top and 4 Ω below. Branch B has 4 Ω on top and 2 Ω below. A 3 µF capacitor joins the midpoint of A to the midpoint of B. Find its charge and which plate is positive.
Take the battery's negative terminal as 0 V.
Branch A carries 12 / 6 = 2 A, so its midpoint is 2 × 4 = 8 V above the bottom.
Branch B also carries 2 A, so its midpoint is 2 × 2 = 4 V above the bottom.
The capacitor sees 8 − 4 = 4 V, so Q = 3 µF × 4 V = 12 µC. The plate joined to branch A is positive.
With no current through the capacitor, the branches are independent voltage dividers, so writing node potentials is faster than loop equations.
Worked example 3: two capacitors in series in one branch
Suppose the 10 V across R₂ in example 1 feeds two capacitors in series instead of one: C₁ = 2 µF and C₂ = 3 µF. Series capacitors carry the same charge. C_eq = (2 × 3)/(2 + 3) = 1.2 µF, so Q = 1.2 × 10 = 12 µC. Then V₁ = 12/2 = 6 V and V₂ = 12/3 = 4 V, which add up to 10 V. The smaller capacitor takes the larger share of the voltage.
Why do these problems trip students up?
Many students reason about circuits as if the battery always pushes out the same current. Engelhardt and Beichner built a 29-question diagnostic test on DC resistive circuits and found that students often held several misconceptions even after instruction. In their interviews, the idea that the battery is a constant source of current was the one used most often. With capacitors that habit is costly, because the current in a capacitor branch falls to zero whatever the battery.
Which mistakes cost marks?
Sending current through the capacitor branch. At steady state it is zero, so a resistor in series with a capacitor has no voltage across it.
Using the battery EMF as the capacitor voltage. Use the potential difference between the capacitor's own two nodes.
Forgetting internal resistance. It takes its share of the voltage, as in example 1.
Mixing up the two time limits. Just after closing, an uncharged capacitor acts as a wire. Long after, it acts as a gap.
Ignoring polarity. In bridge circuits the plate at the higher potential carries +Q.
A practice routine
Day 1: Without notes, write the steady-state rules and the just-after-closing rule.
Day 2: Redo the three examples with new values, and check that series capacitor voltages add up to the branch voltage.
Day 3: Solve three bridge circuits by assigning node potentials.
Day 4: Mix capacitor circuits with plain resistor circuits. Our list of Kirchhoff's law mistakes is a useful checklist.
Day 5: Test yourself on a mixed set without opening the chapter. In their review of study techniques, Dunlosky and colleagues (2013) gave practice testing and spaced practice their highest utility rating.
If the steps of a circuit solution feel like guesswork, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, swap the 4 Ω and 2 Ω in branch B of example 2 and check that the capacitor's charge becomes zero.
