Capacitor question practice
This presentation is the property of its rightful owner.
Sponsored Links
1 / 12

Capacitor Question Practice PowerPoint PPT Presentation


  • 37 Views
  • Uploaded on
  • Presentation posted in: General

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? 

Download Presentation

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Capacitor question practice

Capacitor Question Practice

A2 Physics


Capacitor question practice

Q1

What is the energy held by a

50 000 mF capacitor charged to 12.0 V?

(2 marks)


Capacitor question practice

A1

E = ½ CV2

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


Capacitor question practice

Q2

What is the charge held by a

470 mF capacitor charged to a p.d. of 8.5 V? 

(2 marks)


Capacitor question practice

A2

Q = CV (P)

= 470 x 10-6 F × 8.5 V

= 4.0 × 10-3 C (P)


Capacitor question practice

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)


Capacitor question practice

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)


Capacitor question practice

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)


Capacitor question practice

4a

Time constant = RC 

= 2000 W × 5000 × 10-6 F

= 10 s (P)


Capacitor question practice

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)


Capacitor question practice

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)


Capacitor question practice

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)


  • Login