An uncharged capacitor is connected in series with a battery, a switch and a resistor, as shown in Fig. 6.1.

Question 3

An uncharged capacitor is connected in series with a battery, a switch and a resistor, as shown in Fig. 6.1.

Fig. 6.1

The battery has e.m.f. 9.0 V and negligible internal resistance. The capacitance of the capacitor is 4700 μF.

The switch is closed at time t = 0.

During the time interval t = 0 to t = 4.0 s, the charge passing through the resistor is 22 mC.

(a) (i) Calculate the energy transfer in the battery during the time interval t = 0 to t = 4.0 s. [2]

(ii) Determine, for the capacitor at time t = 4.0 s,

1. the potential difference V across the capacitor, [2]

2. the energy stored in the capacitor. [2]

(b) Suggest why your answers in (a)(i) and (a)(ii) part 2 are different. [1]

Reference: Past Exam Paper – June 2014 Paper 41 & 43 Q6

Solution:

(a) (i)

{We want to find the energy transfer in the battery. So, we use an equation that relates energy to the charge from the battery. We cannot use an equation for capacitors.

e.m.f. E = Work done / Charge

E = energy / Q

Energy = EQ}

Energy = EQ = 9.0 × (22×10^-3)

Energy = 0.20 J                      

(ii)

1.

Capacitance, C = Q / V

Potential difference, V = (22×10^-3) / (4700×10^-6)

p.d. = 4.7 V    

2.

EITHER Energy, E = ½ CV² = ½ × 4700×10^-6 × 4.7² = 5.1×10^-2 J 

OR E = ½ QV = ½ × 22×10^-3 × 4.7 = 5.1×10^-2 J                               

OR E = ½ Q²/C = ½ × (22×10^-3)² / (4700×10^-6) = 5.1×10^-2 J

(b) There is energy lost (as thermal energy) in the resistance / wires / battery / resistor