A 100 μF capacitor initially at 9.00 V discharges through a 20.0 kΩ resistor. Calculate time constant, initial current, voltage and current at 3.00 s, initial energy, and energy dissipated by then. Describe how a plot of lnV against time determines R if C is known, and explain how a voltmeter of 200 kΩ input resistance changes the effective time constant.

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τ = 2.00 s; initial current = 0.450 mA; at 3.00 s V = 2.01 V and I = 0.100 mA. Initial energy = 4.05 mJ and 3.85 mJ has dissipated. With the voltmeter, R_eff = 18.18 kΩ and τ = 1.82 s. Time constant is RC = (20.0 × 10³)(100 × 10⁻⁶) = 2.00 s. Initial current is V₀/R = 9.00/20000 = 0.450 mA. Discharge voltage is V = V₀e^(−t/τ). At 3.00 s, V = 9.00e^(−1.5) = 2.01 V and I = V/R = 0.100 mA. Initial capacitor energy is ½CV₀² = 4.05 mJ. Remaining energy is ½C(2.01)² = 0.202 mJ, so about 3.85 mJ has become thermal energy. Taking logarithms gives lnV = lnV₀ − t/(RC). A best-fit gradient m gives R = −1/(mC), while residual curvature can reveal leakage or changing resistance. The 200 kΩ voltmeter is in parallel with 20.0 kΩ, so R_eff = (20.0×200)/(220) kΩ = 18.18 kΩ and τ = 1.82 s. Ignoring meter loading would underestimate the intended resistor or overestimate capacitance.

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