Circuits - series and parallel
Q6. X and Y are two lamps. X is rated at 12V 36W and Y at 4.5V 2.0W.
(a) Calculate the current in each lamp when it is operated at its correct working voltage.
P = IV
I = P/V
I (lamp X) = 36/12 = 3.0A 
I (lamp Y) = 2.0/4.5 = 0.44A 
(2 marks)
(b) The two lamps are connected into the circuit as shown in the diagram below.

(i) Calculate the pd across R1. 24 - 12 = 12V 
(ii) Calculate the current in R1. current = 3.0 + 0.44 = 3.44A 
(iii) Calculate the resistance of R1. V = IR so R = V/I = 12/3.44 = 3.5 Ω 
(iv) Calculate the pd across R2. 12 - 4.5 = 7.5 V 
(v) Calculate the resistance of R2. V = IR so R = V/I = 7.5/0.44 = 17 Ω 
(5 marks)
(c) The filament of the lamp in X breaks and the lamp no longer conducts. It is observed that the voltmeter reading decreases and lamp Y glows more brightly.
(i) Explain without calculation why the voltmeter reading decreases.
When the bulb blows the resistance of the parallel section increases
therefore it takes a bigger share of the [ptential difference provided by the battery
and the voltage across resistor 1 will decrease, making the voltmeter reading drop.
OR the resistance of the complete circuit increases,
thus the current through resistor 1 will decrease, and as R = V/I so will the potential difference across it.
(2 marks max)
(ii) Explain without calculation why the lamp Y glows more brightly.
The pd across Y (or current through Y) increases
hence the power (rate of energy dissipation) will be greater
making the temperature of the lamp increase
(2 marks max)
(Total 11 marks)
