Q1.

(a) A set of decorative lights consists of a string of lamps. Each lamp is rated at 5.0 V, 0.40 W and is connected in series to a 230 V supply.

Calculate

(i) the number of lamps in the set, so that each lamp operates at the correct rating,

Each lamp needs 5V across it – the mains supplies 230V – so you need 230/5 in series to share it out:

230/5 = 46

(ii) the current in the circuit,

When the lamps have the correct voltage across them they let a current pass through that makes the lamp convert 0.40 J of electrical energy every second (the 0.40W rating indicates that)

P=IV

0.40 = I x 5

so, I = 0.40/5 = 0.080A

Note that you must NOT put 0.08A – that would be putting the answer to only 1SF

(iii) the resistance of each lamp,

V = IR so R = V/I

You can do it for an individual bulb straight off as you have the values:

R = 5.0/0.080 = 62.5

OR you can do it for the whole circuit and then divide by the number of bulbs:

R = 230/(0.080 x 46) = 62.5

OR you can do it from the power equation for a single bulb

P = IV but V = IR,

so P = V²/R and therefore R = V²/P

R = 5.0²/0.40 = 62.5

All of the methods give you the same answer - andy one is valid - but it helps if you add a few words explaining what you are doing!

You should quote the final answer to 2sf as that is the sensitivity of the figures in the question.

R = 63

(iv) the total electrical energy transferred by the set of lights in 2 hours.

The power rating tells you how much energy is transferred in each second, by each bulb. So, in each second 46 bulbs will transfer 46 x 0.40 J of energy.

You are asked for the transfer in 2 hours – so you need to work out how many seconds that would be: 2h = 2 x 60² s

For each bullb energy transfer per second = 46 x 0.40 J

For each bullb energy transfer in two hours= 46 x 0.40 x 2 x 60² J = 1.325 x 10⁵ J

(This should be given to 2sf) = 1.3 x 10⁵ J

(5)

(b) When assembled at the factory, one set of lights inadvertently contains 10 lamps too many. All are connected in series. Assume that the resistance of each lamp is the same as that calculated in part (a) (iii).

(i) Calculate the current in this set of lights when connected to a 230 V supply.

In series therefore add the resistances:

62.5 x (10 + 46)

= 3.5 x 10³

V= IR

so, I = V/R = 230/(3.5 x 10³)

= 0.0657 A

= 0.066 A

(if you use 63 instead of 62.5 you get 0.065A – either answer gets you the mark)

(ii) How would the brightness of each lamp in this set compare with the brightness of each lamp in the correct set?

Each lamp in this set would be dimmer that the corrrct set (as the current through each of them would be lower, due to a smaller p.d. being across them).

(3)
(Total 8 marks)