Two capacitors, a switch and a surprising reading
A student builds a circuit: a 12 V ideal battery, a switch S, and two capacitors in series - C₁ = 4.0 µF and C₂ = 12 µF - with a voltmeter connected across C₂. The voltmeter has finite resistance R_V = 10 MΩ; the capacitors are ideal. Initially both are uncharged. The switch is closed and the student waits 'long enough' for the reading to stabilise. She predicts the voltmeter will read 9.0 V, because the voltage divides in inverse proportion to capacitance. After a long time the voltmeter actually reads 0 V, and she is puzzled.
1. Explain the student's prediction and why it is correct only in the ideal case with no voltmeter.
Read the worked answer
For series capacitors the charge on each is equal: Q = C₁V₁ = C₂V₂, so V₂/V₁ = C₁/C₂. With C₁ = 4 µF and C₂ = 12 µF, V₂ = V·C₁/(C₁+C₂) = 12 × 4/16 = 3.0 V - wait, careful: the larger voltage appears across the smaller capacitance. V₁ = 12 × 12/16 = 9.0 V and V₂ = 12 × 4/16 = 3.0 V.
So the student's number 9.0 V is attached to the wrong capacitor: the ideal reading across C₂ would be 3.0 V, not 9.0 V. Her inverse-proportion reasoning is right, but she applied it to the wrong element. This is the first judgement error to catch.
2. Now include the voltmeter. Explain why the final reading is 0 V and find the time scale over which it decays.
Read the worked answer
The voltmeter is a resistor R_V = 10 MΩ in parallel with C₂. Once the switch is closed and the transient is over, the capacitors are no longer isolated: a DC path exists through the voltmeter. In the steady state no current flows (capacitors block DC), so there is no voltage drop across R_V, hence V₂ → 0.
Physically, the charge originally placed on C₂ leaks through R_V, and the battery maintains the full 12 V across the series combination; in the steady state all 12 V appears across C₁ (which blocks the current) and 0 V across C₂ (shorted by the resistor at DC).
The decay time constant is set by the resistance and the capacitance it discharges. The relevant capacitance seen by R_V is C₂ in parallel with the series combination of C₁ and the (ideal, zero-resistance) battery - effectively C₂ + C₁ = 16 µF, since the battery is a short for signals. So τ ≈ R_V(C₁+C₂) = 10⁷ × 16×10⁻⁶ = 160 s.
After a few τ (≈ 8 minutes) the reading is essentially 0 V. The student 'waited long enough' and indeed saw the steady state - she just misidentified what the steady state should be.
3. The student wants to measure the 3.0 V across C₂ without disturbing it. Propose a fix and justify it with a quantitative criterion.
Read the worked answer
Use a measuring instrument whose input resistance is much larger than the impedance it loads - or better, an electrometer/oscilloscope with 1 GΩ or higher, or measure quickly before discharge.
Criterion: the reading is accurate to within a fraction ε if the leakage time constant is long compared with the measurement time t_m, i.e. R_V(C₁+C₂) ≫ t_m. With R_V = 10 MΩ the time constant is 160 s, so reading within 1 % requires t_m ≪ 1.6 s - feasible only with a fast meter.
Better: use R_V = 1 GΩ (τ = 1.6×10⁴ s) and read at leisure; the loading error is then negligible. Alternatively, buffer the node with an op-amp voltage follower (input resistance ≫ 10¹² Ω).
Justification: the voltmeter must draw negligible charge compared with Q₂ = C₂V₂ = 12 µF × 3 V = 36 µC; a 10 MΩ meter drawing current at 3 V removes 0.3 µA, draining the charge in ~10² s. The fix must reduce that current by orders of magnitude.