Questions on EMF and internal resistance
Q7. 
(a) A steady current of 0.25 A passes through a torch bulb for 6 minutes.
Calculate the charge which flows through the bulb in this time.

Q = 0.25 A
t = 6 mins = 6 x 60s = 360s 
Q = I x
t
Q = 0.25 × 360
Q = 90 C 
(2)
(b) The torch bulb is now connected to a battery of negligible internal resistance. The battery supplies a steady current of 0.25 A for 20 hours.
In this time the energy transferred in the bulb is 9.0 × 104 J. Calculate
(i) the potential difference across the bulb,

V = E/Q
Q = I
t = (0.25 x (20 x 602)
V = E/Q = (9.0 × 104)/(0.25 x (20 x 602) 
V = 5.0V 
(ii) the power of the bulb.
P = IV
P = 0.25 x 5.0
P = 1.3W 
(or you could use P = E/
t)
(3)
(Total 5 marks)