Checking a 10 µF Capacitor: When the Voltmeter Becomes Part of the Circuit
A student is checking whether a box of 10 µF capacitors is within tolerance. She charges one capacitor to 12.0 V and discharges it through a 4.7 MΩ resistor. A digital voltmeter, set to the 20 V range, is connected across the capacitor; its specification gives an input resistance of 10 MΩ. The voltage is recorded every 5 s. The voltage falls from 12.0 V to 6.0 V in 22.2 s. Using τ = RC with R = 4.7 MΩ, the student calculates C = 6.8 µF and concludes that the capacitor is about 32% low and should be rejected. The resistor is marked ±5%, and the voltmeter's reading has an accuracy of ±(0.5% + 2 digits). The student's stopwatch reaction time is about 0.2 s. The capacitor's marked tolerance is ±20%. The capacitor is clean and dry, and the experiment is at room temperature.
1. Calculate the time constant from the measured half-life and the apparent capacitance using the marked 4.7 MΩ resistance. Then calculate the effective resistance of the resistor and voltmeter in parallel, and hence the corrected capacitance.
Read the worked answer
For exponential discharge, the half-life is
[ t_{1/2}=\tau\ln2 ]
so
[ \tau=\frac{22.2}{\ln2}=\frac{22.2}{0.693}\approx32.0\ \text{s} ]
Using only the marked resistor:
[ C_{\text{apparent}}=\frac{\tau}{R}=\frac{32.0}{4.7\times10^6}=6.8\times10^{-6}\ \text{F} ]
But the voltmeter is connected across the capacitor, so the discharge current has two parallel paths: the 4.7 MΩ resistor and the 10 MΩ voltmeter. The effective resistance is
[ R_{\text{eff}}=\frac{(4.7\times10^6)(10\times10^6)}{4.7\times10^6+10\times10^6} \approx3.20\times10^6\ \Omega ]
Corrected capacitance:
[ C=\frac{32.0}{3.20\times10^6}=1.00\times10^{-5}\ \text{F}=10.0\ \mu\text{F} ]
The voltmeter lowers the effective resistance and therefore shortens the time constant for a given capacitance. Using only the marked resistor underestimates the capacitance.
2. Evaluate the student's conclusion that the capacitor should be rejected. Include the marked tolerance, the resistor tolerance and the timing uncertainty.
Read the worked answer
The corrected capacitance is about (10.0\ \mu\text{F}), not (6.8\ \mu\text{F}). The marked tolerance is ±20%, so acceptable values are roughly 8.0–12.0 µF. The corrected value is well inside that range.
The resistor is marked ±5%, so (R_{\text{eff}}) is uncertain by about 5%, giving a capacitance range of roughly 9.5–10.5 µF. The stopwatch uncertainty is about (0.2/22.2\approx0.9%), and the voltmeter accuracy is about 0.5% plus a few digits. These are much smaller than the 20% tolerance.
The student's conclusion is therefore wrong. The apparent 32% deficit is almost exactly the loading effect of the voltmeter: (R_{\text{eff}}/R_{\text{marked}}=3.20/4.7=0.68), so the apparent capacitance is 68% of the true value. This is a systematic instrumental error, not evidence that the capacitor is faulty.
3. Suggest one experimental change that would make the voltmeter loading negligible, and state two precautions needed to keep the discharge clean and reproducible.
Read the worked answer
A suitable change is to measure the voltage with a buffer amplifier or electrometer whose input resistance is much greater than 4.7 MΩ, for example 100 MΩ or more. Alternatively, use a much smaller discharge resistor, such as 100 kΩ, so that the voltmeter's 10 MΩ input is negligible in parallel; the capacitance would then need to be larger to keep a convenient time constant, and a datalogger should be used because the discharge would be faster.
Precautions:
- Clean and dry the capacitor terminals and the insulating bench to avoid leakage currents across the surface, which would add an unknown parallel resistance.
- Use the same initial voltage and polarity each time, and record the voltage with a datalogger or with the same meter on a fixed range, so that offset and reading errors are consistent. Keep the apparatus at room temperature and avoid touching the terminals during discharge.