Q#387: [Current of Electricity > Power dissipated] (Past Exam Paper – June 2010 Paper 11 Q33 & Paper 12 Q34 & Paper 13 Q35)
The resistors P, Q and R in the circuit have equal resistance.
The battery, of negligible internal resistance, supplies a total power of 12W.
What is power dissipated by heating in resistor R?
A 2 W B 3 W C 4 W D 6 W
Solution 387:
Answer: A.
EITHER
Power supplied by battery = VI = 12 W where I is the total current across the circuit and E is the e.m.f. of the battery.
We need to find the (relative) current flowing through each resistor and the potential difference across each of them.
Current flows from the positive terminal of the battery. So, the total current I would flow through resistor P. At the junction, the current splits equally into 2 since the resistances of Q and R are equal. So, the currents through Q and R are each I/2.
From Kirchhoff’s law, the sum of potential difference across any loop is equal to the sum e.m.f of the battery in the circuit. Also, the p.d. across the components in a parallel combination is the same.
The equivalent resistance of the parallel combination of resistors Q and R is half their original resistance. From the potential divider equation, the p.d. across resistor P is twice that across the parallel combination (and thus, across Q or R). Let the e.m.f of the battery be V. The p.d. across P is 2V/3 while the p.d. across Q or R is V/3 (p.d. across P is twice, giving the sum of p.d. in any loop = V).
Power supplied by battery = VI = 12 W
Resistor R: Current = I/2 p.d. = V/3
Power dissipated in resistor R = (I/2) (V/3) = (VI) / 6 = 12 / 6 = 2W
OR
This could alternatively be done by calculations.
Let the resistances of resistors P, Q and R be r each.
Equivalent resistance of parallel combination = [1/r + 1/r]-1 = 0.5r
Total resistance in circuit, rtotal = r + 0.5 = 1.5r = 3r / 2
Total power dissipated in circuit = I2rtotal where I is the total current in the circuit.
Total current I = √(Power / rtotal)
As explained before, the total current I passes through P, but only half of the current passes through Q and R.
Current through P, Ip = √(Power / rtotal) = √(12 / 1.5r) = √(8/r)
Half of this current passes through R.
Current IR through R = Ip/2 = 0.5 √(8/r) = √(2/r)
Power dissipated in resistor R = IR2r = (2/r) r = 2W
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