Questions on charging and discharging

Question 1

Describe the shape of the charge-time graph for a capacitor discharging through a resistor.

Accepted Answer

Exponential decay from the initial (maximum) charge (1); the charge falls quickly at first, then ever more slowly, approaching zero without reaching it (1).

Question 2

Describe how the current changes with time while a capacitor charges through a resistor.

Accepted Answer

The current is a maximum at the instant charging begins (1); it then decreases exponentially towards zero as the capacitor charges (1).

Question 3

State how the current in a capacitor circuit can be found from the charge-time graph.

Accepted Answer

The current is the magnitude of the gradient of the charge-time graph, I = ΔQ/Δt; on discharge that gradient is negative because charge is leaving the plates (1).

Question 4

State what is represented by the area under a current-time graph for a capacitor discharging through a resistor.

Accepted Answer

The total charge transferred (the charge that leaves the capacitor) (1).

Question 5

A capacitor is charged through a resistor from a 5.0 V supply. When charging is complete, state the potential difference across the capacitor and the potential difference across the resistor.

Accepted Answer

The capacitor pd equals the supply pd, 5.0 V (1); the resistor pd is zero, because the current has fallen to zero (1).

Question 6

An initially uncharged capacitor is charged through a resistor. Describe how the potential difference across the resistor varies with time.

Accepted Answer

It starts at its maximum value, equal to the full supply pd (1); it then decreases (exponentially) towards zero as the capacitor pd rises towards the supply pd (1).

Question 7

A capacitor charged to 12 V is discharged through a 1.0 kΩ resistor. Calculate the initial discharge current.

Accepted Answer

I₀ = V/R = 12/1000 (1) I₀ = 0.012 A (12 mA) (1)

Question 8

A capacitor is charged from a 6.0 V supply through a 2.0 kΩ resistor. State the value of the charging current at the start of charging and when the capacitor is fully charged.

Accepted Answer

Initial current I₀ = V/R = 6.0/2000 = 0.0030 A (3.0 mA) (1) Final current = 0 when fully charged (1).

Question 9

A capacitor is charged through a fixed resistor. Describe and explain the shapes of the graphs of potential difference across the capacitor against time and of current against time.

Accepted Answer

The pd across the capacitor rises from zero, rapidly at first, levelling off towards the supply voltage (exponential growth towards a limit) (1). As charge builds up, the pd across the resistor (supply pd minus capacitor pd) falls, so the rate of charging decreases (1). The current starts at V/R and decays exponentially towards zero (1); because the current is driven by the pd across the resistor, which falls as the capacitor charges (1).

Question 10

A 100 μF capacitor is fully charged to 9.0 V. Calculate the charge stored and the energy stored.

Accepted Answer

Q = CV = 100 × 10⁻⁶ × 9.0 = 9.0 × 10⁻⁴ C (1) E = ½CV² = ½ × 100 × 10⁻⁶ × 9.0² (1) E = 4.05 × 10⁻³ J (1)

Question 11

A student discharges a capacitor through a resistor and uses a datalogger to record the charge-time graph. The tangent to the graph at t = 0 has a gradient of −4.0 × 10⁻⁴ C s⁻¹. The initial potential difference across the capacitor is 6.0 V. Determine the resistance of the resistor.

Accepted Answer

The magnitude of the initial gradient is the initial current, I₀ = 4.0 × 10⁻⁴ A (1) R = V₀/I₀ = 6.0/(4.0 × 10⁻⁴) (1) R = 15000 Ω (15 kΩ) (1)

Question 12

An initially uncharged capacitor is fully charged from a 9.0 V supply through a resistor. The area under the current-time graph for the whole charging process is 5.4 mC. Determine the capacitance of the capacitor.

Accepted Answer

The area is the total charge stored, Q = 5.4 × 10⁻³ C (1) C = Q/V = (5.4 × 10⁻³)/9.0 = 6.0 × 10⁻⁴ F (600 μF) (1)

Question 13

A 12 V supply charges a capacitor through a 15 kΩ resistor. At one instant the potential difference across the capacitor is 7.5 V. Calculate the potential difference across the resistor and the charging current at this instant.

Accepted Answer

V_R = 12 − 7.5 = 4.5 V (1) I = V_R/R = 4.5/15000 (1) I = 3.0 × 10⁻⁴ A (0.30 mA) (1)

Question 14

A student is asked how a capacitor charges and writes: “The battery drives charge through the capacitor, from one plate across to the other.”
An uncharged capacitor is charged from a battery through a resistor, and is later discharged through the same resistor. Describe, in terms of electrons, what happens in the circuit during charging and during discharging, and state in each case what passes between the plates.

Accepted Answer

Charging: the battery removes electrons from the plate connected to its positive terminal, leaving that plate with +Q, and delivers electrons through the leads to the plate connected to its negative terminal, which gains −Q (1). The plates gain equal and opposite charge because one current passes round the single loop, supplying one plate at the same rate as it drains the other (1). Discharging: with the battery removed, the pd between the plates drives electrons through the external resistor from the negative plate to the positive plate, until the plates neutralise and the pd is zero (1). In neither case does any charge pass between the plates, so the student is wrong: the dielectric in the gap is an insulator, and all that changes there is the electric field (1).

Question 15

A 470 μF capacitor charged to 10 V is discharged through a 2.2 kΩ resistor. Calculate the initial charge stored and the initial discharge current.

Accepted Answer

Q₀ = CV = 470 × 10⁻⁶ × 10 (1) Q₀ = 4.70 × 10⁻³ C (1) I₀ = V/R = 10/2200 (1) I₀ = 4.55 × 10⁻³ A (4.55 mA) (1)

Question 16

A 1000 μF capacitor is charged to 12 V and then fully discharged through a resistor. Determine the total energy dissipated in the resistor.

Accepted Answer

All the stored energy is transferred to the resistor (1) E = ½CV² = ½ × 1000 × 10⁻⁶ × 12² (1) E = 0.072 J (1)

Question 17

Explain why the current decreases with time as a capacitor discharges through a resistor.

Accepted Answer

As charge leaves the capacitor, the pd across it falls, since V = Q/C (1). The same pd appears across the resistor, and I = V/R (1); a smaller pd drives a smaller current, so the current keeps decreasing as the discharge proceeds (1).

Question 18

A 2200 μF capacitor in an emergency door alarm is charged to 6.0 V. The sounder stops working when the pd across the capacitor has fallen to 3.0 V. Calculate the energy transferred to the sounder circuit between these two pds.

Accepted Answer

E at 6.0 V = ½CV² = ½ × 2200 × 10⁻⁶ × 6.0² = 0.0396 J (1) E at 3.0 V = ½ × 2200 × 10⁻⁶ × 3.0² = 0.0099 J (1) Energy transferred = 0.0396 − 0.0099 = 0.030 J (1)

Question 19

A student is building a circuit in which a 4700 μF capacitor charges from a 12 V supply through a single resistor. Two resistors are available: 100 Ω and 100 kΩ. The student wants the capacitor to charge as quickly as possible, but the charging current must never exceed 50 mA. Deduce which resistor the student should use.

Accepted Answer

The charging current is largest at the first instant, I₀ = V/R (1)
With 100 Ω: I₀ = 12/100 = 0.12 A = 120 mA, which exceeds the 50 mA limit (1)
With 100 kΩ: I₀ = 12/100000 = 1.2 × 10⁻⁴ A = 0.12 mA, which is within the limit (1)
A smaller resistance charges the capacitor faster, but only the 100 kΩ resistor keeps the current below 50 mA, so the student must use 100 kΩ (1).