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)