A capacitor C is charged using a supply of e.m.f. 8.0 V. It is then discharged through a resistor R.

Question 5

A capacitor C is charged using a supply of e.m.f. 8.0 V. It is then discharged through a resistor R.

The circuit is shown in Fig. 5.1.

Fig. 5.1

The variation with time t of the potential difference V across the resistor R during the

discharge of the capacitor is shown in Fig. 5.2.

Fig. 5.2

(a) During the first 1.0 s of the discharge of the capacitor, 0.13 J of energy is transferred to the resistor R.

Show that the capacitance of the capacitor C is 4500 μF. [3]

(b) Some capacitors, each of capacitance 4500 μF with a maximum working voltage of 6 V, are available.

Draw an arrangement of these capacitors that could provide a total capacitance of 4500 μF for use in the circuit of Fig. 5.1. [2]

Reference: Past Exam Paper – June 2008 Paper 4 Q5

Solution:

(a)

{From the graph,}

at t = 1.0 s, V = 2.5 V

Energy of charged capacitor = ½ CV²  

{Note that this formula is for the energy stored in a charged capacitor (not the energy transferred to the resistor).

Initially (at t = 0), energy of charged capacitor = ½ × C × 8² J

At time t = 1 s, energy of charged capacitor = ½ × C × 2.5² J

During this time, the energy is lost by the capacitor is transferred to the resistor.

Energy transferred = 0.13 J = energy of capacitor at t=0 – energy of capacitor at t=1}

0.13 = ½ × C × (8.0² – 2.5²)         

C = 4500 μF                                     

(b)

{When capacitors are connected in series, the equivalent capacitance decreases and when connected in parallel, the equivalent capacitance decreases.

Arranging 2 such capacitors reduces the capacitance to half. So, by connecting another branch of 2 capacitors in parallel, the capacitance doubles, resulting in a capacitance of 400 μF again.

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