Transcript Title
Capacitor Question Practice
A2 Physics
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Q1
What is the energy held by a
50 000 mF capacitor charged to
12.0 V?
(2 marks)
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A1
E = ½ CV2
E = ½ × 50 000 × 10-6 F × (12.0 V)2 (P)
= 3.6 J (P)
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Q2
What is the charge held by a
470 mF capacitor charged to a p.d.
of 8.5 V?
(2 marks)
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A2
Q = CV (P)
= 470 x 10-6 F × 8.5 V
-3
= 4.0 × 10 C (P)
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Q3
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)
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A3
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)
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Q4
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)
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4a
Time constant = RC
= 2000 W × 5000 × 10-6 F
= 10 s (P)
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4b
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)
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4c
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)
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4d
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)
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