Answers to Capacitance

Charge, Voltage and Capacitance 

  1. 	C = 40x10-4F = 4000mF	
  2. 	C = 3x10-5F  = 30mF
  3. 	Q = 7.52x10-4C = 752mC	
  4. 	Pd = 12V 	
  5.	Pd = 9.1V
  6.	I = 3x10-6A = 3mA

  7.	A	(1996 PI  Q17)
  8.	D	(1995 PI  Q17)
  9.	B	(1997 PI  Q16)
 10.	D	(1999 PI  Q18)

 11.i.	Q = 90x10-4C = 9000mC
   ii.	The voltage across the larger capacitor will
 	be less.

	V = Q/C

	The charge is the same in each case but as the value of
	C increases V must decrease.

Energy Stored in a Capacitor (Pages 2-5)

 12.	E = 1.5J
 13.	Pd = 1,500V
 14.	Q = 3x10-3C = 3mC
 15.	E = 3x10-3J = 3mJ
 16.	Pd = 19.1V
 17.	Q = 43.4-3C = 43.4mC
 18.	P = 125W

 19.	  A		(1994 PI  Q17)
 20.	  D		(1995 PI  Q18)
 21.	  C		(1999 PI  Q19)
 22.	  E		(1996 PI  Q18)
 23.i.(A) Q = 0.096C	(1997 PII Q7b)
      (B) E = 288J
   ii.	  I = 48A
 

Charging and Discharging Graphs 

 24.	E	(1992 PI  Q20)
 25.	E	(1995 PI  Q19)
 26.	B	(1993 PI  Q18)
 27.	C	(1994 PI  Q16)

 28.	Position B
 29.	Position A
 30.	
	
 31.	
	
 32.	Imax(charging) = 6x10-4A = 0.6mA
 33.	Vcap(final) = 6V
 34.	Q = 0.012C = 12mC
 35.	
	
 36.	
	
 37.	Imax(discharging) = 3x10-4A = 0.3mA
 38.	
	

 39.	B
 40A.i.	When Vcap reaches 100V the capacitor discharges
	through the neon lamp. This continues until the voltage 
	across the lamp falls to 80V. During this phase the lamp 
	is ON.

    ii.	The capacitor then starts to charge and the voltage across
	it increases until it reaches 100V. During this phase the 
	lamp is OFF.

   iii.	Back to step i above.

   B.
	

   C.	Reduce R or C.

  	


RC Circuit Analysis

 41.	No initial charge on the capacitor.
 42.	No initial voltage across the capacitor. VC = 0V
 43.	VR = 10V
 44.	Imax = 1x10-6A = 1mA
 45.	VC(final) = 10V
 46.	The current in the circuit is determined by the
	voltage across the resistor. As this voltage decreases 
	as the voltage acoss the copacitor increases, during
	charging, the current in the circuit must decrease.
 
 47.	As the charging current decreases the rate at which 
	charge is added to the capacitor also decreases. This 
	reduces the rate VC increases.(DVC = DQ/C)
 
 48.	
	
 49.
	
 50.	Reverse the change in polarity as often as possible.
	This reduces the voltage across the capacitor and
	keeps VR and I as high as possible. ( I = VR/R)
	
 51.	D
 52.	VR = 9.5V
 53.i.	Q = 0.02C	(1999 PII  Q7bi)
   ii.	E = 0.360J	(1999 PII  Q7bii)

 54.a.			(1998 PII  Q6a)
    b.i.  R = 2x105W	(1998 PII  Q6bi)
    b.ii.		(1998 PII  Q6bii)
	
 55.i.				(1998 PII  Q6ci)
   ii.				(1998 PII  Q6cii)
  iii.	C = 2x10-4F = 200mF	(1998 PII  Q6ciii)

 56.a.				(1992 PII  Q6a)
	
    b.	  I = 441.2mA	(1992 PII  Q6b)
    c.i.  E = 40.5mJ	(1992 PII  Q6ci)
    c.ii. Same energy.	(1992 PII  Q6cii)

 57.a.i.  Vsupply = 15V	(1997 PII  Q7ai)
     ii.  Time = 40s	(1997 PII  Q7aii)
    iii.  Vcapacitor = 9V	(1997 PII  Q7aiii)


Capacitors in AC Circuits 

 58.	C	(1996 PI  Q20)
 59.	B	(1998 PI  Q19)

 60.	The ratio Vc/Vs will decrease.	(1995 PI  Q35)
 61.		(1993 PII  Q5b)
 62.		(1992 PI  Q36)
	
General Questions

 63.	C			(1997 PI  Q18)
 64.i.	C = 16mF			(1993 PII  Q5ai)
   ii.	Mean(C) = 17mF		(1993 PII  Q5aii)
	Random error(C) = +-1mF
  iii.				(1993 PII  Q5aiii)