Topic Menu

Cyber-chess a chess instruction site

 

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.

Calculate

(i) the potential difference between the points A and B in the circuit,

RAB = 5.0s s
V = IR
= 5.0 × 0.50 = 2.5Vs

(ii) the internal resistance of the battery,

Vr = 12 – (2.5 + 5.0) s
= 4.5 V s
r = Vr/I
= 4.5/0.5 = 9.0 ss

(iii) the total energy supplied by the battery in 2.0 s,

Energy = EMF x It
12 x 0.5 x 2 = 12Js

(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
Ratio of energy = 4.5/12 = 0.375
= 38% s

(Max 7)
(Total 7 marks)