OCR A Physics · Module 6: Particles and medical physics
Capacitors: practice questions with answers
6 exam-style questions on capacitors. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Unit prefixes2 marks
A 470 μF capacitor is charged to 12 V. Calculate the charge stored, in mC.
Q = mC
Show the answer and working
Answer Q = 5.64 mC
Q = CV = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C = 5.64 mC.
- Got 0.00564 mC? That is the charge in coulombs (5.64 × 10⁻³ C). Multiply by 1000 to give mC: 5.64 mC.
- Got 5640 mC? You converted μF using 10⁻³. 1 μF = 10⁻⁶ F, so Q = CV = 470 × 10⁻⁶ × 12 = 5.64 mC.
- Unit prefixes2 marks
A 2200 μF capacitor is charged to 9.0 V. Calculate the energy stored, in mJ.
E = mJ
Show the answer and working
Answer E = 89.1 mJ
E = ½CV² = ½ × 2200 × 10⁻⁶ × 9.0² = 0.0891 J = 89.1 mJ.
- Got 178 mJ? You left out the ½. E = ½CV² = ½ × 2200 × 10⁻⁶ × 9.0² = 89.1 mJ.
- Got 9.9 mJ? You did not square the voltage. E = ½CV² = ½ × 2200 × 10⁻⁶ × 81 = 89.1 mJ.
- Using the right equation3 marks
A 100 μF capacitor charged to 6.0 V discharges through a 47 kΩ resistor. Calculate the p.d. across the capacitor after 3.0 s.
V = V
Show the answer and working
Answer V = 3.17 V
RC = 47 × 10³ × 100 × 10⁻⁶ = 4.7 s. V = V₀e^(-t/RC) = 6.0 × e^(-3.0/4.7) = 3.17 V.
- Got 2.83 V? You used the charging equation. For discharge V = V₀e^(-t/RC), with RC = 4.7 s: V = 6.0e^(-3.0/4.7) = 3.17 V.
- Got 4.51 × 10⁻⁶ V? You multiplied t by RC instead of dividing. V = V₀e^(-t/RC) = 6.0e^(-3.0/4.7) = 3.17 V.
- Using the right equation3 marks
A 10 μF capacitor discharges through a 220 kΩ resistor. Calculate the time taken for its p.d. to fall to 20% of its initial value.
t = s
Show the answer and working
Answer t = 3.54 s
RC = 2.2 s. 0.20 = e^(-t/RC), so t = RC ln 5 = 2.2 × 1.609 = 3.54 s.
- Got 11 s? You used t = 5RC. The decay is exponential: 0.20 = e^(-t/RC), so t = RC ln 5 = 2.2 × 1.609 = 3.54 s.
- Got 0.44 s? You divided RC by 5. Solve 0.20 = e^(-t/RC): t = RC ln 5 = 3.54 s.
- Using the right equation1 mark
Two capacitors of 4.0 μF and 6.0 μF are connected in series. Which is the correct total capacitance?
- A 2.4 μF
- B 10 μF
- C 24 μF
Show the answer and working
Answer A: 2.4 μF
1/C = 1/4.0 + 1/6.0 = 5/12, so C = 2.4 μF, smaller than either capacitor.
- Why not B? Capacitances add in parallel, not in series.
- Why not C? This is the product, not the series combination 1/C = 1/4 + 1/6.
- Using the right equation1 mark
The resistance in a capacitor discharge circuit is doubled. What happens to the time constant?
- A It halves.
- B It doubles.
- C It stays the same.
Show the answer and working
Answer B: It doubles.
τ = RC, so doubling R doubles τ and the discharge takes twice as long.
- Why not A? The time constant RC is proportional to R, so doubling R makes the discharge slower.
- Why not C? The time constant depends on R, so it changes.