capacitor question practice
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Capacitor Question Practice. A2 Physics. Q1. What is the energy held by a 50 000 m F capacitor charged to 12.0 V? (2 marks) . A1. E = ½ CV 2 E = ½ × 50 000 × 10 -6 F × (12.0 V) 2 ( P ) = 3.6 J ( P ). Q2. What is the charge held by a 470 m F capacitor charged to a p.d. of 8.5 V? - PowerPoint PPT PresentationTRANSCRIPT
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Capacitor Question PracticeA2 Physics
Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved.
Q1What is the energy held by a
50 000 mF capacitor charged to 12.0 V?
(2 marks)
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A1E = ½ CV2
E = ½ × 50 000 × 10-6 F × (12.0 V)2 (P) = 3.6 J (P)
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Q2What is the charge held by a 470 mF capacitor charged to a
p.d. of 8.5 V? (2 marks)
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A2Q = CV (P)
= 470 x 10-6 F × 8.5 V = 4.0 × 10-3 C (P)
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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)
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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)
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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)
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4a
Time constant = RC = 2000 W × 5000 × 10-6 F
= 10 s (P)
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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)
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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)
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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)