A capacitor discharges through 220 kΩ with (time in s, voltage in V) readings (0, 8), (10, 5.84), (20, 4.26), (30, 3.11), (40, 2.27). (a) Estimate time constant and capacitance. (b) Find initial charge and energy. (c) Find initial current and resistor power. (d) Find energy fraction at 40 s. (e) Explain voltmeter loading and a semilog test.

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τ ≈31.8 s; C ≈145 μF; Q₀ ≈1.16 mC; U₀ ≈4.64 mJ; I₀ = 36.4 μA; P₀ = 0.291 mW; energy fraction ≈0.0805. From V = V₀e^(−t/τ), use 5.84/8.0 at 10 s: τ = −10/ln(5.84/8.0) = 31.8 s. C = τ/R = 31.8/220000 = 1.45 × 10⁻⁴ F = 145 μF. Initial charge CV = 1.16 × 10⁻³ C and energy ½CV² = 4.64 × 10⁻³ J. Initial current V/R = 8.0/220000 = 36.4 μA. Initial power V²/R = 2.91 × 10⁻⁴ W. At 40 s, voltage fraction = 2.27/8.0 = 0.28375. Since U ∝V², energy fraction = 0.28375² = 0.0805. A finite-resistance voltmeter adds a parallel path and reduces τ. A straight ln V against t plot with gradient −1/τ tests single-exponential decay.

Capacitor discharge and stored field energy

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Capacitor discharge

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Capacitor discharge — supplied values
Time / s	Voltage / V
0	8
10	5.84
20	4.26
30	3.11
40	2.27