A capacitor discharges through a 110 kΩ resistor. Its voltage readings (time in s, voltage in V) are (0, 9), (12, 4.5), (24, 2.25), (36, 1.125). Assume V = V₀ exp[−t/(RC)] and that the voltmeter draws negligible current. (a) Determine RC and capacitance. (b) Find the fraction of initial stored energy remaining after 24 s. (c) Explain what changes if a voltmeter of finite resistance is connected across the capacitor.

Question

Accepted Answer

Time constant 17.31 s; capacitance 157.4 μF; energy fraction 1/16; finite meter resistance makes discharge faster. Voltage halves every 12 s, so 1/2 = exp[−12/(RC)]. Taking natural logarithms gives RC = 12/ln 2 = 17.31 s. A voltage half-time is not the same as the time constant. Convert the discharge resistance to 110000 Ω. Then C = (RC)/R = 17.31/110000 = 0.0001574 F = 157.4 μF. All three halving intervals support the same exponential model. After 24 s the voltage is one-quarter of its original value. Since stored energy is ½CV² for fixed capacitance, the energy fraction is (1/4)² = 1/16, not 1/4. A finite-resistance voltmeter forms an additional parallel discharge path. The effective resistance is smaller than the stated resistor alone, reducing the time constant and speeding the discharge. Using the resistor's value without correcting this loading would underestimate the capacitance.

Finding capacitance from a discharge 2

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Capacitor voltage readings

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Capacitor voltage readings — supplied values
Time / s	Voltage / V
0	9
12	4.5
24	2.25
36	1.125