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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Capacitor Question Practice A2 Physics

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Q1 What is the energy held by a 50 000 F capacitor charged to 12.0 V? (2 marks)

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A1 E = ½ CV 2 E = ½ × 50 000 × 10 -6 F × (12.0 V) 2 ( ) = 3.6 J ( )

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Q2 What is the charge held by a 470 F capacitor charged to a p.d. of 8.5 V? (2 marks)

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. A2 Q = CV ( ) = 470 x 10 -6 F × 8.5 V = 4.0 × 10 -3 C ( )

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 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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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 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) ( ) C = 9.38 × 10 -6 F = 9.38 mF ( )

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. Q4 A 5000 F capacitor is charged to 12.0 V and discharged through a 2000 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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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 4a Time constant = RC = 2000 × 5000 × 10 -6 F = 10 s ( )

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 4b V = V 0 e –t/RC V = 12.0 × e – 13 /10 ( ) V = 12.0 × e – 1.3 = 12.0 × 0.273 = 3.3 volts ( )

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 4c V = V 0 e –t/RC V V 0 = 0.5 = e –t(half)/RC ln(0.5) = - t 1/2 /RC ln2 = t 1/2 /RC( ) (The log of a reciprocal is the negative of that for the original number) t 1/2 = 0.693 × RC = 0.693 × 10 = 6.93 s ( )

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Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 4d V = V 0 e –t/RC V V 0 = e –t/RC ln V - ln Vo = -t/RC ( ) (When you divide two numbers, you subtract their logs) 0.693 – 2.485 = - t/10 ln2 - ln12 = - t/10 ( ) -t/10 = -1.792 t/10 = 1.792 t = 1.792 × 10 = 17.9 s ( )

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