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Energy and Charge Calculation for Capacitors in Circuit Analysis

This document presents a series of physics practice questions focused on capacitors, covering energy stored, charge calculation, and discharge characteristics. The exercises challenge students to apply formulas such as E = ½ CV² and Q = CV to determine key parameters of circuit behavior. Problem-solving involves calculations with different capacitances and voltages, resulting in practical applications of theoretical concepts. It is aimed at A2-level physics students seeking to strengthen their understanding of capacitors in electrical circuits.

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Energy and Charge Calculation for Capacitors in Circuit Analysis

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  1. Capacitor Question Practice A2 Physics

  2. Q1 What is the energy held by a 50 000 mF capacitor charged to 12.0 V? (2 marks)

  3. A1 E = ½ CV2 E = ½ × 50 000 × 10-6 F × (12.0 V)2 (P) = 3.6 J (P)

  4. Q2 What is the charge held by a 470 mF capacitor charged to a p.d. of 8.5 V?  (2 marks)

  5. A2 Q = CV (P) = 470 x 10-6 F × 8.5 V = 4.0 × 10-3 C (P)

  6. 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)

  7. 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)

  8. 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)

  9. 4a Time constant = RC  = 2000 W × 5000 × 10-6 F = 10 s (P)

  10. 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)

  11. 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)

  12. 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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