Help with Paper 1 question

[thecsstudent]

How do we go about finding the answer for this paper 1 exercise?

So far I have thought of this:

V = E - Ir (where V is p.d. across the motor)

E/t = C/t *V => E = C*V => V = E/C = 9000/450 = 20 V

So 20 = E - Ir

So I am guessing the answer is 24 because E needs to be greater than 20.

Edited March 14, 2016 by thecsstudent

[Vioh]

@thecsstudent the way that you think about the problem is correct. However, you don't need to "guess", you can actually calculate exactly what the answer must be. The way that the diagram is drawn is actually a hint, because from the diagram, you can see that the circuit consists of an EMF connected in series with a motor and a resistor.

We have the formula: P = E/t = V*I = V*C/t => E = VC (where E is the total energy dissipated in the circuit, V is the EMF, and C is the charge that passes through the circuit). The total energy dissipated is E = 9000 + 1800 = 10800. Since the circuit is connected in series, so the charge passing through the motor (which is 450C) is also the charge passing through the whole circuit, so C = 450. Hence,

EMF = V = E/C = 10800 / 450 = 24V. So yes, D is the correct answer.

[Physics Tutor]

Vioh is correct, yet I prefer a more straightforward way to think about this (remember you do not have much time for each question at the exam).

Energy is conserved in the circuit, so if 9000+1800J are dissipated, that means 10800 J have been provided by the cell.

The emf is energy provided by the cell to the circuit per unit charge (EMF=W/Q), so 10800/450 = 24V.

Answer D