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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

A common mistake is not to change the time into seconds – extracting data onto the right hand side and checking it is in the correct unit avoids this pitfall!

(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,

From the data sheet we have the relationship between charge, energy and voltage

– in this case it is V not we use, but in the same relationship:

V = E/Q

The question gives us E (9.0 × 104 J) – we have to calculate Q in the same way as above – but remember t MUST be in seconds

Q = It = (0.25 x (20 x 602)

V = E/Q = (9.0 × 104)/(0.25 x (20 x 602)

V = 5.0V (note that again you get a mark for converting the hours into seconds – it is important to ensure time is always in seconds!)

(ii) the power of the bulb.

P = IV

P = 0.25 x 5.0

P = 1.3W (1.25W rounds up)

(or you could use P = E/t)

(3)

(Total 5 marks)