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A capacitor in a circuit with a battery does not charge instantly. The current starts large and dies away, and the capacitor's voltage creeps up towards the battery's. JEE Advanced questions on this use a small set of ideas: an exponential, a time constant, and what the capacitor looks like at the start and at the end. This post derives the formulas from Kirchhoff's loop rule, works through two circuits and lists the mistakes that cost marks.
Which exams test RC circuits?
The JEE (Advanced) 2026 brochure lists "RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources" under electromagnetic induction. The JEE (Main) 2026 syllabus covers capacitors and the LCR series circuit in AC, but does not name RC charging. So this is mainly a JEE Advanced and olympiad topic. Check the current year's syllabus for your exam.
Where do the formulas come from?
Take a battery of emf E, a resistor R and an uncharged capacitor C in series, and close the switch at t = 0. Walking round the loop (Kirchhoff's loop rule):
E − IR − q/C = 0, with I = dq/dt.
This is a first-order differential equation. Separating variables and integrating from q = 0 at t = 0 gives
q = CE(1 − e^(−t/RC))
I = (E/R) e^(−t/RC)
V_C = E(1 − e^(−t/RC))
For discharge through R from charge Q₀, the loop rule is q/C + R dq/dt = 0, so q = Q₀ e^(−t/RC) and the current falls the same way. If the separation-of-variables step is rusty, our post on differential equations for JEE covers it.
The product RC has units of time: ohms times farads is seconds. It is called the time constant, τ.
What does the time constant tell you?
At t = τ, the factor e^(−1) ≈ 0.37. So while charging, the capacitor reaches 1 − 0.37 ≈ 63% of its final charge after one time constant. While discharging, it falls to about 37%. After 5τ, it is within 1% of the end value, which is why "five time constants" is a common rule for "fully charged".
Two shortcuts for the start and the end:
At t = 0, an uncharged capacitor has no voltage across it, so it behaves like a plain wire.
As t → ∞, no current flows into it, so it behaves like a break in the circuit.
Worked example 1: a simple charging circuit
E = 10 V, R = 100 kΩ, C = 20 µF.
τ = RC = 100 × 10³ × 20 × 10⁻⁶ = 2.0 s.
Initial current: E/R = 10 / 10⁵ = 100 µA.
Final charge: CE = 20 × 10⁻⁶ × 10 = 200 µC.
At t = 2.0 s (one τ): q = 200 × (1 − e^(−1)) ≈ 200 × 0.632 ≈ 126 µC, so V_C ≈ 6.3 V.
Time to reach 90% of full charge: 1 − e^(−t/τ) = 0.9 gives t = τ ln 10 ≈ 2.0 × 2.303 ≈ 4.6 s. Time to reach half charge: τ ln 2 ≈ 1.39 s.
Where does the energy go? The battery moves 200 µC through 10 V, supplying QE = 2.0 mJ. The capacitor ends up with ½CE² = ½ × 20 × 10⁻⁶ × 100 = 1.0 mJ. The other 1.0 mJ is dissipated in the resistor. That half-and-half split does not depend on R.
Worked example 2: two resistors, JEE Advanced style
A 12 V battery is connected in series with R₁ = 6 kΩ and R₂ = 3 kΩ. A 10 µF capacitor, initially uncharged, is connected across R₂. The switch closes at t = 0.
Final voltage: long after, the capacitor is a gap, so the current is 12 / 9 kΩ and the voltage across R₂ is 12 × 3/9 = 4 V. That is the capacitor's final voltage.
Time constant: use the resistance the capacitor sees, with the battery replaced by a wire. R₁ and R₂ are then in parallel: 6 × 3 / (6 + 3) = 2 kΩ. So τ = 2 × 10³ × 10 × 10⁻⁶ = 0.020 s, or 20 ms.
Result: V_C = 4(1 − e^(−t/0.020)) volts.
After the battery is disconnected, the capacitor discharges through R₂ only (R₁ is now in an open branch), so the new time constant is 3 kΩ × 10 µF = 30 ms.
Why do students struggle with RC circuits?
The calculus is rarely the issue. Coppens, Van den Bossche and De Cock (2017) gave a questionnaire on simple RC circuits to over 150 engineering students in Belgium after lectures and labs. Students rarely recognised the circuits as filters, and many failed to apply Kirchhoff's laws and Ohm's law correctly to reach the answer. In other words, the basic circuit rules broke down once a capacitor was present. Our post on Kirchhoff's laws and sign conventions is good preparation.
Which mistakes cost marks?
Using the wrong resistance in τ. Use the resistance seen from the capacitor's terminals, with batteries replaced by wires.
Swapping 63% and 37%. Charging reaches 63%. Discharging falls to 37%.
Units. kΩ × µF = 10³ × 10⁻⁶ = 10⁻³ s. A common slip gives seconds instead of milliseconds.
Assuming energy is conserved. Half the battery's energy goes into the resistor during charging.
If you understand each formula but cannot see how the pieces fit in a new circuit, Fermi's AI tutor can work out what you already know and build a plan for the session, so you sit down to a clear next step.
A practice routine
Day 1: Derive the charging and discharging equations from the loop rule, without looking.
Day 2: Five single-resistor problems: τ, charge at a given time, time to reach a given fraction.
Day 3: Three circuits with two or three resistors, using "wire, then gap" for the end values and the seen resistance for τ.
Day 4: Two energy problems and a timed mixed set. Revise capacitor networks first with our guide to capacitors and dielectrics.
Tonight, redo Worked example 2 with R₁ = R₂ = 3 kΩ and check that τ = 15 ms and the final voltage is 6 V.
