Questions on EMF and internal resistance
Q3.
A battery of e.m.f. 12 V and internal resistance r is connected in a circuit with three resistors each having a resistance of 10
as shown. A current of 0.50A flows through the battery.
![](Q2.JPG)
Calculate
(i) the potential difference between the points A and B in the circuit,
RAB = 5.0
![s](../../../../graphics/symbols/nuclides/ticksmall.png)
V = IR
= 5.0 × 0.50 = 2.5V![s](../../../../graphics/symbols/nuclides/ticksmall.png)
(ii) the internal resistance of the battery,
Vr = 12 – (2.5 + 5.0) ![s](../../../../graphics/symbols/nuclides/ticksmall.png)
= 4.5 V ![s](../../../../graphics/symbols/nuclides/ticksmall.png)
r = Vr/I
= 4.5/0.5 = 9.0 ![s](../../../../graphics/symbols/omega_red.png)
![s](../../../../graphics/symbols/nuclides/ticksmall.png)
(iii) the total energy supplied by the battery in 2.0 s,
Energy = EMF x It
12 x 0.5 x 2 = 12J![s](../../../../graphics/symbols/nuclides/ticksmall.png)
(iv) the percentage of the energy supplied by the battery that is dissipated within the battery.
Energy dissipated in battery = VrIt
= 4.5 x 0.5 x 2 = 4.5J ![s](../../../../graphics/symbols/nuclides/ticksmall.png)
Ratio of energy = 4.5/12 = 0.375
= 38% ![s](../../../../graphics/symbols/nuclides/ticksmall.png)
(Max 7)
(Total 7 marks)