Last saved by: Joseph Dixon

1.	Calculation of charge
	6000 V ´ 20 ´ 10^–6 F    (1)
= 0.12 C    (1)	2
	Energy stored in capacitor
	     (1)
= 360 J    (1)	2
	Resistance
= 150 W    (1)	1
	Time to discharge capacitor
Time =  /their Q    (1)
= 0.0030 s / 3.0 ´ 10^–3 s [e.c.f.]    (1)	2
	Reason
Time is longer because the rate of discharge decreases/ current decreases 
with time    (1)	1
[8]

2.	Correct circuit	1
	Straight line through origin 
	Q_MAX = 900 µC and sensible scale 
	V_MAX = 4.5 V and sensible scale	3
	Appropriate formula: ½ CV² or ½QV 
2 × 10^–³ J	2
[6]

3.	The circuit shown is used to charge a capacitor.

The graph shows the charge stored on the capacitor whilst it is being charged.

	On the same axes, sketch as accurately as you can a graph of current against time.  Label the current axis with an appropriate scale.
Label current axis     (1)
Current at t = 0 within range 30 – 45 mA     (1)
Current graph right shape     (1)
Exponential decay     (1)
(4 marks)
The power supply is 3 V.  Calculate the resistance of the charging circuit.
Resistance = 3 V / 40 mA     (1)
	= 75 kW     (1)
Resistance  = Allow 66 kW ®100 kW
(2 marks)
[Total 6 marks]

4.	Slope of graph:
Capacitance
Shaded area of graph:
Energy/work done	2
Energy stored 3.1 J:
CV^2/2
= 100 ´ 10^–6 × 250^2/2 [formula + correct substitution]
(= 3.125) = 3.1 J  [Must have previous mark]	2

Power from cell, and minimum time for cell to recharge capacitor:
Cell power 	= 1.5 V × 0.20 A
	= 0.30 W [allow 3/10 W here]
Time	= 3.1 J/0.30 W(e.c.f.)
	= 10 s	3
[7]

5.	Relationship between current and charge:
	Current is the rate of flow of charge/rate of change of charge OR current is charge per second
OR  I = Q/t (with or without d or D) but with symbols defined	(1)
1
Explanation:
	Since I is constant, Q o on capacitor (= It) increases at a steady rate OR 
charge flows at a constant rate	(1)
Since V µ Q, V also increases at a steady rate	(1)
OR  
V = Q/C = It/C	(1)
and V = (I/C) × t compared with y = (m) × x
2
Determination of current, using graph:
Use of Q = CV | Attempt to get grad | (1)
Use of I = Q/t | Use of I = C × grad | (1)
= 1.1 mA | 1.1 mA | (1)
3
Explanation:
Decrease [If increase, 0/3]	(1)
As capacitor charges, V_R decreases	(1)
R must decrease because I = V_R/R OR R must decrease to prevent I 
falling	(1)
3
Second graph:
Line added to graph showing:
Any curve getting less steep with time [from origin; no maximum]	(1)
And with same initial gradient as original straight line	(1)
2
[11]

6.	Estimation of charge delivered:
Charge 	=	area under graph (1)
	=	a number of squares × correct calculation for charge of 
	one square i.e. correct attempt at area e.g. single triangle (1)
	=	(3.5 to 4.8) × 10^–3 C (A s, mA s) (1)
[Limit = triangle from 41 mA ® 300 s]
OR
Charge 	=	average current × time (1)
	=	(something between 10 and 20 mA) × 300 s (1)
	=	(3.5 to 4.8) × 10^–3 C (1)	3
[But Q = It ® 0/3, e.g. 41 mA × 300 s]
Estimation of capacitance
C	=	calculated charge/9.0V	time constant »100 s (1)
	=	390 to 533 mF	C = 100 s/220 k.W = 450 mF (1)	2
[5]

Beverley High School