Capacitors in DC Steady State
At DC steady state, replace an ideal capacitor by an open circuit, solve the resistive network for its terminal voltage, then use Q = CV and U = one-half CV squared.
Why this shows up in the exam
Capacitor charge in resistor networks · Energy stored after switching settles · Initial-versus-final circuit comparison
Learn the idea
After a long time on a DC source, an ideal capacitor is open to current but can hold voltage and energy. During charging, current moves charge onto the plates. Once the capacitor voltage stops changing, no conduction current passes through the ideal capacitor branch, although the charged capacitor still affects node voltages.
🧠 Memory hook: Long-time DC: open branch, but voltage and charge may remain.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- I_C = C dV_C/dt = 0 — ideal capacitor at constant steady voltage
- Q = C V_C — stored charge after finding capacitor voltage
- U = (1/2) C V_C² — stored electrostatic energy
How to approach it
- 1Decide initial, transient, or final state
- 2Open the capacitor for final DC
- 3Solve V_C, then calculate Q or U
Common slip-ups that cost marks
- •Shorting the capacitor at long time
- •Using the battery emf without solving node voltage
- •Applying steady state immediately after switching
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 2 ohm resistor. Find the circuit current.
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