Questions on the time constant and exponential decay

Question 1

State the equation for the time constant of a capacitor-resistor circuit and explain what the time constant represents for a discharging capacitor.

Accepted Answer

τ = RC (1); the time for the charge (or pd, or current) to fall to 1/e (about 37%) of its initial value (1).

Question 2

A 100 μF capacitor discharges through a 10 kΩ resistor. Calculate the time constant of the circuit.

Accepted Answer

τ = RC = 10000 × 100 × 10⁻⁶ (1) τ = 1.0 s (1)

Question 3

Write down the equation for the charge remaining on a capacitor at time t as it discharges through a resistor.

Accepted Answer

Q = Q₀e^−t/RC, where Q₀ is the initial charge (1).

Question 4

A capacitor charges through a resistor. State the percentage of its final charge that it has gained one time constant after charging begins.

Accepted Answer

63% (the fraction 1 − 1/e) (1).

Question 5

A 47 μF capacitor discharges through a 2.2 MΩ resistor. Calculate the time constant of the circuit.

Accepted Answer

τ = RC = 2.2 × 10⁶ × 47 × 10⁻⁶ (1) τ = 103 s (1)

Question 6

A 470 μF capacitor discharges through a 2.2 kΩ resistor. Calculate the time constant and state the fraction of the initial charge that remains after one time constant.

Accepted Answer

τ = RC = 2200 × 470 × 10⁻⁶ (1) τ = 1.03 s (1) After one time constant, Q/Q₀ = 1/e ≈ 0.37 (37%) (1).

Question 7

A 100 μF capacitor is charged to 12 V and then discharged through a 47 kΩ resistor. Calculate the potential difference across the capacitor 5.0 s after the discharge begins.

Accepted Answer

τ = RC = 47000 × 100 × 10⁻⁶ = 4.7 s (1) V = V₀e^−t/RC = 12 × e^−5.0/4.7 (1) V = 4.14 V (1)

Question 8

For the circuit in the previous question, calculate the time taken for the potential difference to halve.

Accepted Answer

T_½ = 0.69RC = 0.69 × 4.7 (1) T_½ = 3.24 s (1)

Question 9

A timing circuit requires a time constant of 2.0 s using a 220 μF capacitor. Calculate the resistance required.

Accepted Answer

R = τ/C = 2.0/(220 × 10⁻⁶) (1) R = 9.1 × 10³ Ω (9.1 kΩ) (1)

Question 10

In required practical 9 a student discharges a capacitor through a resistor while logging the pd across it. The pd falls from 6.0 V to 2.2 V in 15 s. Show that the time constant of the circuit is about 15 s.

Accepted Answer

V = V₀e^−t/τ, so t/τ = ln(V₀/V) (1) ln(6.0/2.2) = 1.00 (1) τ = 15/1.00 = 15.0 s ≈ 15 s (1)

Question 11

A capacitor discharges through a 33 kΩ resistor. The pd across it halves every 4.6 s. Determine the capacitance of the capacitor.

Accepted Answer

T_½ = 0.69RC (1) C = T_½/(0.69R) = 4.6/(0.69 × 33000) (1) C = 2.0 × 10⁻⁴ F (200 μF) (1)

Question 12

A capacitor discharges through a resistor with a time constant of 3.0 s. Determine the time taken for the potential difference across it to fall from 20 V to 5.0 V.

Accepted Answer

5.0 = 20e^−t/3.0, so e^−t/3.0 = 0.25 (1) t = −3.0 × ln(0.25) (1) t = 4.16 s (1)

Question 13

A 470 μF capacitor is charged from a 9.0 V supply through a 10 kΩ resistor. Calculate the potential difference across the capacitor 10 s after charging begins.

Accepted Answer

τ = RC = 10000 × 470 × 10⁻⁶ = 4.7 s (1) V = V₀(1 − e^−t/RC) = 9.0 × (1 − e^−10/4.7) (1) V = 7.93 V (1)

Question 14

A student plots ln(V) against time for a discharging capacitor. Explain how the time constant can be determined from this graph.

Accepted Answer

Taking logs of V = V₀e^−t/RC gives ln V = ln V₀ − t/RC (1); this is a straight line with gradient −1/RC (1); so the time constant RC is the negative reciprocal of the gradient (1).

Question 15

A student uses required practical 9 to test a capacitor labelled 470 μF with a tolerance of ±20%. The capacitor is discharged through a 10 kΩ resistor, and the graph of ln(V) against t is a straight line of gradient −0.185 s⁻¹. Deduce whether the capacitor is within its stated tolerance.

Accepted Answer

The gradient is −1/RC, so RC = 1/0.185 = 5.41 s (1)
C = RC/R = 5.41/10000 = 5.41 × 10⁻⁴ F = 541 μF (1)
The tolerance allows 376 μF to 564 μF (1)
541 μF lies inside this range, so the capacitor is within its stated tolerance (1).

Question 16

A capacitor discharges through a resistor. Show that the energy stored falls to about 14% of its initial value after one time constant.

Accepted Answer

E = ½Q²/C, so E is proportional to Q² (1) After one time constant Q has fallen to Q₀/e, so E falls to E₀/e² (1) 1/e² = 0.135, which is about 14% (1)

Question 17

The manual for a high-voltage supply states that its reservoir capacitor must be left to discharge until less than 1.0% of the initial charge remains before the case is opened. The discharge time constant is 4.0 s. Determine the minimum waiting time.

Accepted Answer

Q/Q₀ = 0.010 = e^−t/τ (1) t = −4.0 × ln(0.010) = 4.0 × ln(100) (1) t = 18.4 s (about 4.6 time constants) (1)

Question 18

Describe an experiment to determine the capacitance of an unmarked capacitor by discharging it through a resistor of known resistance. Include the measurements you would take and explain how you would analyse the data to obtain the capacitance.

Accepted Answer

Charge the capacitor from a supply, then discharge it through the known resistor with a voltmeter or datalogger connected across the capacitor (or resistor) (1). Record the pd at regular time intervals, using a stopwatch (or the datalogger's clock) started as the discharge begins (1). Plot ln V against t (1); a straight line confirms that the decay is exponential (1). The gradient of the line equals −1/RC (1); so C = −1/(R × gradient), using the known resistance (1).

Question 19

A 100 μF capacitor carrying a charge of 600 μC begins to discharge through a 20 kΩ resistor at t = 0. A student models the discharge step by step from ΔQ/Δt = −Q/CR, using a time step Δt = 0.50 s and holding the rate at its start-of-step value across each step. Calculate the charge left by the model at the end of each of the four steps to t = 2.0 s, compare the final value with the charge given by Q = Q₀e^−t/CR, and explain the difference.

Accepted Answer

CR = 20000 × 100 × 10⁻⁶ = 2.0 s, so ΔQ = −(Q/CR)Δt = −Q × 0.50/2.0 = −0.25Q, leaving 0.75Q at the end of each step (1)
Step 1: ΔQ = −150.0 μC, Q = 450.0 μC; step 2: ΔQ = −112.5 μC, Q = 337.5 μC (1)
Step 3: ΔQ = −84.4 μC, Q = 253.1 μC; step 4: ΔQ = −63.3 μC, Q = 189.8 μC (1)
The exponential gives Q = 600 × e^−2.0/2.0 = 220.7 μC (1)
The model's 189.8 μC is below the exponential's 220.7 μC, low by about 14% (1)
because the rate is held at its start-of-step value, which is the largest it takes during the step, so every step removes slightly too much charge; a smaller Δt reduces the difference (1)