🔬Carry out the required practical: capacitor discharge
Discharging a Capacitor
When a charged capacitor (initial charge Q₀, voltage V₀) is connected to a resistor R, it discharges exponentially. The current decreases as the charge decreases because there is less p.d. to drive the current.
Q = Q₀ e^(−t/RC)
V = V₀ e^(−t/RC)
I = I₀ e^(−t/RC)
Q₀ = initial charge (C)
V₀ = initial voltage (V)
I₀ = initial current = V₀/R (A)
t = time (s), R = resistance (Ω), C = capacitance (F)
All three quantities (Q, V, I) decay with the same exponential time constant RC. The negative exponent means the quantity gets smaller over time. After one time constant: Q = Q₀/e ≈ 0.37Q₀ (37% of initial value remains).
Time Constant and Half-Life
Time constant (τ): τ = RC. The time for Q (or V or I) to fall to 1/e (≈ 36.8%) of its initial value during discharge.
τ = RC
Units: Ω × F = (V/A) × (C/V) = C/A = s ✓
Half-life of discharge:
t½ = RC ln 2 = 0.693 RC
Time to fall to any fraction f of initial value:
t = −RC ln(f)
Time elapsed
Fraction of Q₀ remaining
0
100%
τ = RC
37%
2τ
14%
3τ
5%
5τ
0.7% (essentially discharged)
A larger RC means slower discharge (more time to reach a given fraction). The half-life t½ = 0.693RC — similar to radioactive decay, each half-life reduces Q by a factor of 2.
Charging a Capacitor
When an uncharged capacitor is connected to a DC supply ε through a resistor R, it charges exponentially. Initial current is ε/R (all voltage across R); as charge builds up, more voltage is across the capacitor and less drives the current.
Q = Q₀ (1 − e^(−t/RC)) — charge builds up
V = V₀ (1 − e^(−t/RC)) — voltage builds up
I = I₀ e^(−t/RC) — current decays
Q₀ = Cε (final charge when fully charged)
V₀ = ε (final voltage = EMF)
I₀ = ε/R (initial current)
During charging: Q and V increase exponentially towards their maximum values. Current decreases exponentially (same decay as discharge). After one time constant τ = RC, Q has reached 63% of its final value. After 5τ, the capacitor is considered fully charged (99.3%).
During charging, the voltage across the resistor = ε − V_C. This means the resistor voltage decays while the capacitor voltage rises — they always sum to ε (KVL).
Log Graphs for Exponential Decay
Taking the natural log of the discharge equation:
Q = Q₀ e^(−t/RC)
ln Q = ln Q₀ − t/(RC)
Comparing to y = mx + c:
y = ln Q, x = t
gradient = −1/(RC)
y-intercept = ln Q₀
A graph of ln Q vs t (or ln V vs t) should be a straight line with gradient −1/(RC). This is used in the required practical to verify the exponential nature of discharge and to measure τ = RC.
If the graph of ln Q vs t is straight, it confirms the decay is exponential. Any curvature suggests the decay is not purely exponential (e.g. the capacitor is not ideal, or R varies).
A 100 μF capacitor charged to 10 V is discharged through a 5 kΩ resistor. Calculate: (a) the time constant; (b) the charge remaining after 2.5 s; (c) the voltage after 2.5 s.
3. During the charging of a capacitor, which quantity DECAYS exponentially?
During charging, Q and V across the capacitor increase exponentially. Only the current I = I₀e^(−t/RC) decays — it starts high (maximum when capacitor is empty) and falls to zero as the capacitor reaches full voltage.
4. A student plots ln V vs t for a discharging capacitor. The graph is a straight line. What does this confirm?
If V = V₀e^(−t/RC), then ln V = ln V₀ − t/RC. This is linear in t, so a straight ln V vs t graph confirms the exponential nature of the decay.
5. Calculate the half-life (in seconds) for a capacitor circuit with RC = 4.0 s.
1. A 1000 μF capacitor is charged to 9.0 V and then discharged through a 2.0 kΩ resistor. (a) Find the charge after 4.0 s. (b) Find the energy remaining after 4.0 s. (c) What percentage of the initial energy remains?
τ = RC = 2000 × 1000×10⁻⁶ = 2.0 s. (a) Q₀ = CV = 1000×10⁻⁶ × 9.0 = 9.0×10⁻³ C = 9.0 mC. Q = Q₀e^(−4.0/2.0) = 9.0×10⁻³ × e^(−2) = 9.0×10⁻³ × 0.1353 = 1.22×10⁻³ C = 1.22 mC. (b) V = V₀e^(−2) = 9.0 × 0.1353 = 1.218 V. W = ½CV² = ½ × 1000×10⁻⁶ × 1.218² = 5×10⁻⁴ × 1.483 = 7.42×10⁻⁴ J = 0.742 mJ. (c) Initial energy W₀ = ½CV₀² = ½ × 1000×10⁻⁶ × 81 = 4.05×10⁻² J = 40.5 mJ. % remaining = 0.742/40.5 × 100 = 1.83%. Note: W = ½CV² ∝ V² ∝ e^(−2t/RC), so energy decays twice as fast as voltage (e^(−2t/τ) vs e^(−t/τ)).
2. A student measures the voltage across a discharging capacitor at regular intervals: t=0: 8.0 V; t=10s: 5.9 V; t=20s: 4.3 V; t=30s: 3.2 V; t=40s: 2.4 V. By plotting ln V vs t, determine RC and the capacitance if R = 50 kΩ.
Calculate ln V: t=0: ln(8.0)=2.08; t=10: ln(5.9)=1.775; t=20: ln(4.3)=1.459; t=30: ln(3.2)=1.163; t=40: ln(2.4)=0.875. Gradient = Δ(ln V)/Δt = (0.875 − 2.08)/(40 − 0) = −1.205/40 = −0.03013 s⁻¹. Since gradient = −1/RC: RC = 1/0.03013 = 33.2 s. C = RC/R = 33.2/(50×10³) = 6.64×10⁻⁴ F = 664 μF. (Nearest standard value would be 680 μF.)
3. A capacitor smoothing circuit uses a 2200 μF capacitor to smooth a rectified AC supply at 50 Hz. The load resistance is 100 Ω. Estimate the time constant and explain whether this is sufficient for good smoothing.
τ = RC = 100 × 2200×10⁻⁶ = 0.22 s. A full-wave rectified supply at 50 Hz has a ripple frequency of 100 Hz, so the period between voltage peaks is T = 1/100 = 0.01 s = 10 ms. For good smoothing, the time constant should be much greater than the ripple period: τ >> T. Here τ = 220 ms >> T = 10 ms (τ/T = 22). The capacitor will barely discharge between peaks (it discharges to e^(−10/220) ≈ 0.956 = 95.6% of its peak value before recharging). This gives very good smoothing — the ripple voltage will be very small compared to the DC output voltage. A smaller capacitor (smaller τ) would give more ripple; a larger capacitor gives smoother DC.
Objective: Investigate how the voltage across a capacitor varies with time during discharge, and determine the time constant RC.
Equipment
Capacitor (e.g. 1000 μF electrolytic)
Resistor (e.g. 10 kΩ)
DC power supply (e.g. 10 V)
Voltmeter or datalogger with voltage probe
Switch, stopwatch, connecting leads
Method
Charge the capacitor fully by connecting it to the supply (switch to charge position). Check voltmeter reads the supply voltage.
Disconnect the supply (or flip switch to discharge position) to connect the capacitor to the resistor. Start the stopwatch simultaneously.
Record V every 10 s (or use a datalogger for automatic recording). Record until V < 10% of initial value.
Plot V vs t. Draw a smooth exponential curve.
Plot ln V vs t. Draw a best-fit straight line. Gradient = −1/RC → find R and verify against the labelled resistor value.
Safety
Ensure electrolytic capacitors are connected with correct polarity. Do not exceed the capacitor's voltage rating. Discharge the capacitor fully before handling (short the terminals through a resistor first).
Sources of Uncertainty
Source
Reduction
Reaction time starting stopwatch
Use datalogger; start recording before switching
Voltmeter drawing current
Use digital voltmeter with high input resistance (≥ 10 MΩ)
Capacitance tolerance
Electrolytic capacitors can be ±20% — measure C separately