1 / 12

Capacitor Question Practice

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? 

rusk
Télécharger la présentation

Capacitor Question Practice

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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)

More Related