When a capacitor discharges through a resistor, the voltage falls exponentially. Divide the target voltage by the starting voltage, then use the natural logarithm to find the time. The fraction after one time constant is about 0.37, so reaching one quarter takes more than one time constant.
The 2027 H2 Physics 9478 syllabus covers the variation of quantities during capacitor discharge. This example finds the time for the voltage to fall to one quarter and distinguishes that from the time for stored energy to fall to one quarter.
- Convert resistance and capacitance to SI units.
- Calculate the time constant RC.
- Write the fraction remaining.
- Use the logarithm and check whether the time is plausible.
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Write the voltage ratio
For a capacitor discharging through a fixed resistor, V/V₀ = exp(−t/RC). Taking natural logarithms gives t = −RC ln(V/V₀). A fraction between zero and one has a negative logarithm, so the time is positive.
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Worked example | Find when the voltage falls to one quarter
A 100 μF capacitor discharges through a 20.0 kΩ resistor. The time constant is (100 × 10⁻⁶ F)(20.0 × 10³ Ω) = 2.00 s. If the voltage falls from 8.0 V to 2.0 V, V/V₀ = 0.25. The time is −2.00 ln(0.25) ≈ 2.77 s.
Check against one time constant. At 2.00 s, about 37% of the voltage remains. The capacitor needs additional time to reach 25%.
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Change the target to one quarter of the energy
With capacitance fixed, stored energy is proportional to voltage squared. One quarter of the original energy therefore corresponds to one half of the original voltage. Use V/V₀ = 0.50 to obtain t = 2.00 ln 2 ≈ 1.39 s.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Does charge reach one quarter at the same time as voltage?
Yes, for a fixed capacitance. Charge is proportional to voltage through Q = CV.
What if the question asks when 75% of the voltage has been lost?
Use 25% as the fraction remaining. That gives the same 2.77 s as the worked example.