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Capacitor Questions

Q8. A capacitor of capacitance 330 μF is charged to a potential difference of 9.0V. It is then discharged through a resistor of resistance 470 kΩ.

Calculate

(a) the energy stored by the capacitor when it is fully charged,

Energy stored = ½QV

But Q = CV so

Energy stored = ½CV2

Energy stored = ½ x 330 x 10-6 x 9.02 

Energy stored = 1.34 x 10-2 J 

(2 marks)

(b) the time constant of the discharging circuit,

time constant = RC

time constant = 470 x 103 x 330 x 10-6 = 155

time constant = 160 s 

(1 mark)

(c) the p.d. across the capacitor 60s after the discharge has begun.

V = V 0  e –t/RC

V = 9.0  e –60/155

V = 9.0  e –0.387

V = 9.0 x 0.679

V = 6.1 volts

(3 marks)

(Total 6 marks)