A capacitor is charged to a potential difference of 15 V and then connected in series with a switch, a resistor of resistance 12 kΩ and a sensitive ammeter, as shown in Fig. 5.1.

Question 2

(a) State one function of capacitors in simple circuits. [1]

(b) A capacitor is charged to a potential difference of 15 V and then connected in series with a switch, a resistor of resistance 12 kΩ and a sensitive ammeter, as shown in Fig. 5.1.

Fig. 5.1

The switch is closed and the variation with time t of the current I in the circuit is shown in

Fig. 5.2.

Fig. 5.2

(i) State the relation between the current in a circuit and the charge that passes a

point in the circuit. [1]

(ii) The area below the graph line of Fig. 5.2 represents charge.

Use Fig. 5.2 to determine the initial charge stored in the capacitor. [4]

(iii) Initially, the potential difference across the capacitor was 15 V.

Calculate the capacitance of the capacitor. [2]

(c) The capacitor in (b) discharges one half of its initial energy. Calculate the new potential difference across the capacitor. [3]

Reference: Past Exam Paper – November 2007 Paper 4 Q5

Solution:

(a)

Example:

They separate charges

They store energy

For smoothing circuit

(b)

(i) Charge = current × time                            

(ii)

{Since Charge = current × time, the area under the current-time graph gives the charge.

To obtain the (total) initial charged stored in the capacitor, we need to consider the area under the whole graph, not only the initial part of the graph.

This can be done by either counting the squares or using the area of vertical strips to obtain the total area under the graph.}

Area under graph = 21.2 cm²  (allow ± 0.5cm²)                                 

1.0 cm² represents (0.125×10^-3 × 1.25 =) 156 μC

{1 cm² = 1 cm on y-axis × 1 cm on x-axis

On y-axis, 4 cm represents 0.5 mA

So, 1 cm represents 0.5 / 4 = 1.25 mA = 0.125×10^-3 A

On x-axis, 4 cm represents 5 s

So, 1 cm represents 5 / 4 = 1.25 s}

So, charge = 3300 μC

(iii)

Capacitance = Q / V = (3300×10^-6) / 15 = 220 μF

(c)

EITHER (stored) energy = ½ CV²     OR energy = ½ QV and C = Q/V

{As the capacitor discharges, the amount of charge Q also changes along with the potential difference across the capacitor.

The capacitor (when the p.d. is 15V) discharges half of its initial energy. So, the energy of the capacitor with the new p.d. V has half on the initial energy. In other words, twice this stored energy is equal to the initial stored energy.

Initial stored energy (when p.d. = 15 V) = 2 × energy stored at new p.d. V

Note that the capacitance of the capacitor remains the same.}

½ × C × 15² = 2 × [½ × C × V²]        

V = 10.6 V