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9702 · 19.2

Energy stored in a capacitor — practice questions

Practice and worked examples for 9702 Energy stored in a capacitor. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 220 μFcapacitorischargedtoapotentialdifferenceof12V\mu F capacitor is charged to a potential difference of 12 V. Calculate the energy stored. If it discharges through a 10 k\Omega resistor, what is the initial discharge current and the time constant of the circuit?

Show solution outline
  1. Energy Stored (E): Given: C=220 μF=220×10−6 FC = 220 \, \mu F = 220 \times 10^{-6} \, F, V=12 VV = 12 \, V Formula: E=12CV2E = \frac{1}{2}CV^2 E=12×(220×10−6)×(12)2E = \frac{1}{2} \times (220 \times 10^{-6}) \times (12)^2 E=12×220×10−6×144E = \frac{1}{2} \times 220 \times 10^{-6} \times 144 E=15840×10−6 J=0.01584 JE = 15840 \times 10^{-6} \, J = 0.01584 \, J E≈0.016 JE \approx 0.016 \, J (to 2 s.f.)
  2. Initial Discharge Current (I\textsubscript0): At the start of discharge, the capacitor voltage is V0=12 VV_0 = 12 \, V. The initial current through the resistor is I0=V0RI_0 = \frac{V_0}{R} (from Ohm's Law). Given: R=10 kΩ=10×103 ΩR = 10 \, k\Omega = 10 \times 10^3 \, \Omega I0=1210×103=1.2×10−3 A=1.2 mAI_0 = \frac{12}{10 \times 10^3} = 1.2 \times 10^{-3} \, A = 1.2 \, mA
  3. Time Constant (τ\tau): Formula: τ=RC\tau = RC τ=(10×103 Ω)×(220×10−6 F)\tau = (10 \times 10^3 \, \Omega) \times (220 \times 10^{-6} \, F) τ=2.2 s\tau = 2.2 \, s

Worked example 2

A 200 μF200\,\mu\text{F} capacitor is charged to 12 V12\,\text{V}. Calculate the energy stored using W=12CV2W = \frac{1}{2}CV^2.

Show solution outline

C=200×10−6 FC = 200\times 10^{-6}\,\text{F} W=12CV2=12(200×10−6)(12)2=1.4×10−2 JW = \frac{1}{2}CV^2 = \frac{1}{2}(200\times 10^{-6})(12)^2 = 1.4\times 10^{-2}\,\text{J} (2 s.f.)

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