Capacitor Question Practice

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Capacitor Question PracticeA2 Physics

http://www.cyberphysics.co.uk/


Q1What is the energy held by a

50 000 mF capacitor charged to 12.0 V?

(2 marks)



A1E = ½ CV2

E = ½ × 50 000 × 10-6 F × (12.0 V)2 (P) = 3.6 J (P)



Q2What is the charge held by a 470 mF capacitor charged to a

p.d. of 8.5 V? (2 marks)



A2Q = CV (P)

= 470 x 10-6 F × 8.5 V = 4.0 × 10-3 C (P)



Q3A capacitor is connected to a 12V power supply by a reed switch operating at 400 Hz.

The ammeter reads 45 mA. What is the capacitance of

the capacitor?(2 marks)



A3Q = It but f = 1/t

so Q = I/f

C = Q/V = I ¸ (Vf)C = 0.045 A ¸ (400 Hz × 12.0

V) (P)C = 9.38 × 10-6 F = 9.38 mF (P)



Q4A 5000 mF capacitor is charged to 12.0 V and

discharged through a 2000 W resistor.(a) What is the time constant? (1 mark)(b) What is the voltage after 13 s? (2 marks)(c) What is the half-life of the decay? (2

marks)(d) How long would it take the capacitor to

discharge to 2.0 V (3 marks)(8 marks total)



4a

Time constant = RC = 2000 W × 5000 × 10-6 F

= 10 s (P)



4bV = V0 e –t/RC

V = 12.0 × e – 13 /10 (P)V = 12.0 × e – 1.3

= 12.0 × 0.273 = 3.3 volts (P)



4cV = V0 e –t/RC

V ¸ V0 = 0.5 = e –t(half)/RC ln(0.5) = - t1/2 /RCln2 = t1/2 /RC (P)

(The log of a reciprocal is the negative of that for the original number)

t1/2 = 0.693 × RC = 0.693 × 10 = 6.93 s (P)



4dV = V0 e –t/RC

V ¸ V0 = e –t/RC

ln V - ln Vo = -t/RC (P)(When you divide two numbers,

you subtract their logs)0.693 – 2.485 = - t/10

ln2 - ln12 = - t/10 (P)-t/10 = -1.792

t/10 = 1.792t = 1.792 × 10 = 17.9 s (P)


Capacitor Question Practice

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