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Capacitor Question Practice

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Capacitor Question Practice

A2 Physics

What is the energy held by a

50 000 mF capacitor charged to 12.0 V?

(2 marks)

E = ½ CV2

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

What is the charge held by a

470 mF capacitor charged to a p.d. of 8.5 V?

(2 marks)

Q = CV (P)

= 470 x 10-6 F × 8.5 V

= 4.0 × 10-3 C (P)

A 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)

Q = 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)

A 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)

Time constant = RC

= 2000 W × 5000 × 10-6 F

= 10 s (P)

V = 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)

V = V0 e –t/RC

V ¸ V0 = 0.5 = e –t(half)/RC

ln(0.5) = - t1/2 /RC

ln2 = 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)

V = 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 = 1.792

t = 1.792 × 10 = 17.9 s (P)