Topic Menu

Cyber-chess a chess instruction site
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.

The battery has an emf of 24V and negligible internal resistance. The resistors, R1 and R2 are chosen so that the lamps are operating at their correct working voltage.

To tackle this mark in the voltages across the bulbs - the ones given to you as the operating voltages. You can then tackle the parallel arrangement with bulb X. Bulb Y and resistor 2 will have the same drop as bulb X. The potential drop acroos resistor 1 and the parallel arrangement is the full voltage from the battery.

You should also mark on the currents you worked out in part a.

(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)