# 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. Capacitor Question Practice A2 Physics

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)

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 (  )

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)

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 (  )

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)

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 (  )

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)

Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved. 4a Time constant = RC = 2000  × 5000 × 10 -6 F = 10 s (  )

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 (  )

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 (  )

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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