A Level Physics 19.3 Discharging a Capacitor MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 19.3 of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level topic: how a charged capacitor discharges exponentially through a resistor, the time constant RC, and how to calculate the charge, potential difference and current at any time. All 30 questions are original practice questions with worked answers.
Formula recap
- Discharge: Q = Q0 e−t/RC, V = V0 e−t/RC, I = I0 e−t/RC
- Time constant: τ = RC (unit: ohm × farad = second)
- e−1 = 0.37 and e−2 = 0.14, so after one time constant 37% remains and after two time constants 14% remains
- Half-time for the charge: t1/2 = 0.69 RC
- Charging (the mirror image): Q = Q0 (1 − e−t/RC), so after one time constant the capacitor has reached 63% of its final charge
See the full A Level Physics Formula Sheet for every equation in one place.
Exponential Discharge
Q1. A fully charged capacitor discharges through a fixed resistor. How do the charge Q on the capacitor, the potential difference V across it and the current I in the circuit change with time?
- (A) All three decrease linearly to zero
- (B) Q and V decrease exponentially, but I stays constant
- (C) Q, V and I all decrease exponentially with the same time constant
- (D) Q stays constant while V and I decrease
Show Answer
Answer: C – V = Q/C, so V is proportional to Q, and I = V/R, so I is proportional to V. All three quantities therefore follow the same exponential decay curve with time constant RC.
Q2. In the equation Q = Q0 e−t/RC for capacitor discharge, what do the symbols Q0 and the product RC represent?
- (A) Q0 is the initial charge on the capacitor, and RC is a constant with the unit of time
- (B) Q0 is the charge remaining after time t, and RC is the final charge
- (C) Q0 is the resistance, and RC is the capacitance
- (D) Q0 is the charge on one electron, and RC is the current
Show Answer
Answer: A – Q0 is the charge at t = 0 and Q is the charge remaining at time t. The product RC is the time constant: ohm × farad = (V/A) × (C/V) = C/A = second.
Q3. A capacitor is first charged from a battery and then connected across a resistor to discharge. The discharge current in the resistor is
- (A) in the same direction as the charging current
- (B) zero, because the battery has been removed
- (C) alternating, changing direction many times per second
- (D) in the opposite direction to the charging current
Show Answer
Answer: D – During charging, conventional current flows onto the positive plate. During discharge, conventional current flows away from the positive plate, through the external resistor, to the negative plate, so the current is reversed.
Q4. The charge on a discharging capacitor does NOT fall in a straight-line (linear) graph because
- (A) the resistance of the resistor steadily increases as it warms up
- (B) the rate of loss of charge is proportional to the charge remaining, so the current gets smaller as the charge falls
- (C) the capacitance decreases as the charge falls
- (D) charge can only leave the plates in complete coulombs
Show Answer
Answer: B – The discharge current is I = V/R = Q/(RC). As Q falls, the current falls in proportion, so the charge drains quickly at first and then ever more slowly. A rate of change proportional to the quantity itself always gives exponential decay.
Q5. The discharge of a capacitor through a resistor follows exactly the same mathematics as
- (A) radioactive decay of a nuclide
- (B) motion with uniform acceleration
- (C) simple harmonic motion
- (D) a steady current in a series circuit
Show Answer
Answer: A – In radioactive decay the activity is proportional to the number of undecayed nuclei remaining, just as the discharge current is proportional to the charge remaining. Both processes are exponential and both have a constant half-time.
Q6. Which description matches the charge-time graph for a capacitor discharging through a resistor?
- (A) A straight line falling from Q0 to zero
- (B) A curve that rises from zero and levels off at Q0
- (C) A curve starting at Q0 that falls steeply at first, then more and more gradually, approaching the time axis
- (D) A horizontal line at Q0
Show Answer
Answer: C – The graph is an exponential decay curve. Its gradient (equal to −I) is largest at the start and tends to zero as the charge runs out, so the curve flattens towards the time axis.
Q7. During discharge, the current in the circuit is at its maximum
- (A) at the end of the discharge, just before it stops
- (B) at t = 0, the moment discharge begins
- (C) after exactly one time constant
- (D) after exactly one half-time
Show Answer
Answer: B – The initial current is I0 = V0/R, when the capacitor p.d. is greatest. From then on the current decays exponentially: I = I0 e−t/RC.
Q8. According to the ideal exponential model, after a very long time the charge on a discharging capacitor
- (A) becomes exactly zero after one time constant
- (B) becomes exactly zero after five time constants
- (C) becomes negative, because the plates over-discharge
- (D) tends towards zero but never quite reaches it
Show Answer
Answer: D – An exponential curve approaches zero asymptotically. In practice the remaining charge becomes too small to measure after about five time constants, but mathematically it never reaches exactly zero.
Q9. At any instant during discharge, the potential difference across the capacitor is equal to
- (A) the potential difference across the resistor at the same instant
- (B) the initial potential difference V0, which it keeps throughout
- (C) zero, because the battery has been disconnected
- (D) the product Q0R
Show Answer
Answer: A – The capacitor and resistor form a single loop, so the p.d. across the resistor (IR) must equal the p.d. across the capacitor (Q/C) at every instant. Both fall together exponentially.
Q10. The charge on a discharging capacitor falls to one half of its value in a certain time. In the next equal time interval, the charge will
- (A) fall to zero
- (B) fall by the same amount again, losing the same number of coulombs
- (C) halve again, falling to one quarter of its original value
- (D) stay at one half, because the fastest discharge is over
Show Answer
Answer: C – Exponential decay removes the same FRACTION of the remaining quantity in each equal time interval, not the same amount. So the charge goes Q0 → (1/2)Q0 → (1/4)Q0 in successive half-times.
The Time Constant
Q11. The time constant of a capacitor-resistor discharge circuit is given by
- (A) R/C
- (B) RC
- (C) R + C
- (D) 1/(RC)
Show Answer
Answer: B – The time constant τ = RC. It appears in the exponent of the decay equation Q = Q0 e−t/RC.
Q12. The unit of the product RC is the second. This is because ohm × farad is equivalent to
- (A) volt
- (B) ampere
- (C) joule
- (D) (V/A) × (C/V) = C/A, and one ampere is one coulomb per second
Show Answer
Answer: D – Resistance in ohms is V/A and capacitance in farads is C/V, so RC has unit (V/A)(C/V) = C/A. Since A = C/s, C/A = C/(C/s) = s, a time.
Q13. A capacitor discharges through a resistor. After a time equal to one time constant, the fraction of the original charge remaining on the capacitor is
- (A) 0.37
- (B) 0.50
- (C) 0.63
- (D) 0.14
Show Answer
Answer: A – At t = RC, Q = Q0 e−1 = 0.37 Q0. The same 0.37 factor applies to V and I after one time constant. (0.63 is the fraction LOST, not the fraction remaining.)
Q14. After a time equal to two time constants, the fraction of the original charge remaining is
- (A) 0.74
- (B) 0.25
- (C) 0.14
- (D) 0.37
Show Answer
Answer: C – At t = 2RC, Q = Q0 e−2 = 0.14 Q0. Note that 0.14 is roughly 0.37 × 0.37: each time constant multiplies the remaining charge by 0.37.
Q15. The half-time t1/2 of the charge on a discharging capacitor (the time for the charge to halve) is related to the time constant by
- (A) t1/2 = RC
- (B) t1/2 = 2 RC
- (C) t1/2 = RC / e
- (D) t1/2 = 0.69 RC
Show Answer
Answer: D – Setting e−t/RC = 1/2 gives t = RC ln 2 = 0.69 RC. The charge halves every 0.69 RC, which is shorter than one full time constant.
Q16. To make a capacitor discharge more slowly through a resistor, you should
- (A) decrease both R and C
- (B) increase R, increase C, or increase both
- (C) charge the capacitor to a higher initial voltage
- (D) use a resistor of lower resistance so the charge can escape easily
Show Answer
Answer: B – The rate of discharge is set by the time constant τ = RC. Increasing R reduces the current for a given p.d., and increasing C means more charge must flow for the same drop in p.d. Changing the initial voltage changes Q0 but not the time constant.
Q17. The time constant can be read from a charge-time discharge graph as
- (A) the time taken for the charge to fall to 0.37 of its initial value
- (B) the time taken for the charge to fall to 0.50 of its initial value
- (C) the time taken for the charge to reach zero
- (D) the time taken for the current to reach its maximum value
Show Answer
Answer: A – By definition, after one time constant the charge is Q0 e−1 = 0.37 Q0. The 0.50 point gives the half-time, 0.69 RC, instead.
Q18. In a discharge circuit, the resistance R is doubled while the capacitance C is kept the same. The time constant
- (A) is unchanged
- (B) is halved
- (C) is doubled
- (D) is quadrupled
Show Answer
Answer: C – τ = RC is directly proportional to R, so doubling R doubles the time constant and the discharge takes twice as long.
Q19. After five time constants, the charge remaining on a discharging capacitor is approximately
- (A) 37% of the original charge
- (B) 14% of the original charge
- (C) 5% of the original charge
- (D) less than 1% of the original charge
Show Answer
Answer: D – At t = 5RC the fraction remaining is e−5 = 0.0067, about 0.7%. This is why a capacitor is treated as practically fully discharged after about 5 RC.
Q20. A capacitor with initial charge 100 μC discharges through a resistor. The charge is found to have fallen to 37 μC after 4.0 s. The time constant of the circuit is
- (A) 2.0 s
- (B) 4.0 s
- (C) 0.69 s
- (D) 8.0 s
Show Answer
Answer: B – 37 μC is 0.37 of the initial 100 μC, and the charge always falls to 0.37 of its value after one time constant, so RC = 4.0 s.
Discharging Calculations & Uses
Q21. A 1000 μF capacitor discharges through a 10 kΩ resistor. The time constant of the circuit is
- (A) 10 s
- (B) 1.0 s
- (C) 100 s
- (D) 0.10 s
Show Answer
Answer: A – τ = RC = (10 × 103 Ω) × (1000 × 10−6 F) = 104 × 10−3 = 10 s.
Q22. A capacitor is charged to 12 V and then discharges through a resistor. After one time constant, the potential difference across the capacitor is
- (A) 7.6 V
- (B) 6.0 V
- (C) 4.4 V
- (D) 0 V
Show Answer
Answer: C – After one time constant the p.d. falls to 0.37 of its initial value: V = 0.37 × 12 = 4.44 V ≈ 4.4 V.
Q23. When a capacitor starts to discharge, the current is 2.0 mA. The current after two time constants is
- (A) 0.74 mA
- (B) 0.28 mA
- (C) 1.0 mA
- (D) 0.14 mA
Show Answer
Answer: B – The current decays with the same exponential factor as the charge: I = I0 e−2 = 0.14 × 2.0 mA = 0.28 mA.
Q24. A discharge circuit has a time constant of 2.0 s and uses a 500 μF capacitor. The resistance in the circuit is
- (A) 250 Ω
- (B) 1.0 kΩ
- (C) 40 Ω
- (D) 4.0 kΩ
Show Answer
Answer: D – R = τ/C = 2.0 / (500 × 10−6) = 2.0 / (5.0 × 10−4) = 4000 Ω = 4.0 kΩ.
Q25. A timing circuit needs a time constant of 3.0 s using a 6.0 kΩ resistor. The capacitance required is
- (A) 50 μF
- (B) 2000 μF
- (C) 500 μF
- (D) 18 000 μF
Show Answer
Answer: C – C = τ/R = 3.0 / (6.0 × 103) = 5.0 × 10−4 F = 500 μF.
Q26. An uncharged capacitor is charged through a resistor from a battery of fixed voltage. After one time constant, the charge on the capacitor has reached
- (A) 63% of its final value
- (B) 37% of its final value
- (C) 50% of its final value
- (D) 100% of its final value
Show Answer
Answer: A – Charging follows Q = Q0(1 − e−t/RC). At t = RC, Q = Q0(1 − 0.37) = 0.63 Q0, the mirror image of the discharge curve.
Q27. A capacitor starts with a charge of 80 μC. The half-time for the discharge is 2.0 s. The charge remaining after 6.0 s is
- (A) 20 μC
- (B) 27 μC
- (C) 5.0 μC
- (D) 10 μC
Show Answer
Answer: D – 6.0 s is three half-times. The charge halves three times: 80 → 40 → 20 → 10 μC.
Q28. A capacitor with initial charge 200 μC discharges with a time constant of 5.0 s. The charge remaining after 10 s is
- (A) 74 μC
- (B) 28 μC
- (C) 100 μC
- (D) 7.4 μC
Show Answer
Answer: B – 10 s is two time constants, so Q = Q0 e−2 = 0.14 × 200 μC = 28 μC. (74 μC would be the value after one time constant.)
Q29. Camera flash units and defibrillators both use a charged capacitor, rather than a battery alone, because the capacitor can
- (A) store charge forever without any leakage
- (B) deliver a steady current for a very long time
- (C) release its stored energy in a very short time when discharged through a low resistance
- (D) produce a voltage higher than any battery
Show Answer
Answer: C – With a small resistance, the time constant RC is tiny, so the stored energy is dumped in a brief, powerful pulse. A battery alone cannot deliver energy this quickly because of its internal resistance.
Q30. In a flashing-light timer, a capacitor discharges through a resistor and the light flashes once per discharge cycle. To double the time between flashes, keeping the same capacitor, you should
- (A) double the resistance, doubling the time constant RC
- (B) halve the resistance
- (C) double the charging voltage
- (D) quarter the resistance
Show Answer
Answer: A – The timing is set by τ = RC. Doubling R doubles τ, so the discharge takes twice as long. Doubling the voltage only doubles the starting charge; it does not change the time constant or the timing.
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All questions on this page are original practice questions written for Quaidianpoint.com. They are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.