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Capacitor Questions

Q4. A charged capacitor of capacitance 50 mF is connected across the terminals of a voltmeter of resistance 200 kΩ . When time t = 0, the reading on the voltmeter is 20.0 V.

Calculate

(a) the charge on the capacitor at t = 0,

C = 50 mF

V = 20.0 V

Q = CV = 50 x 10-6 x 20

Q = 1000 x 10-6

Q = 1.0 mC

(1 mark)

(b) the reading on the voltmeter at t = 20s,

Time constant = CR = 50 x 10-6 x 200 x 103 = 10 s

Q = Q o e -t/CR

Q = 1.0 × 10 –3 × e –2

Q = 0.14 × 10 –3 C

V = Q/C = 2.8 V

(2 marks)

(c) the time which must elapse, from t = 0, before 75% of the energy which was stored in the capacitor at t = 0 has been dissipated.

final energy stored = 0.25 (1/2 Q o 2/C)

Q = 0.5 Q o = Q o e –t/10

t = 6.9 s

(3 marks)

(Total 6 marks)