Answer to Question #259376 in Electricity and Magnetism for yuu

Question

A wire of resistance 5  is connected to a battery whose emf is 2 V and whose internal resistance is 1.0 . In 2.0 min (a) how much energy is transferred from chemical to electric form? (b) How much energy appears in the wire as thermal energy? (c) Account for the differences between (a) and (b). 

Answer 1

a) The transfered energy

$U = P \ t =\frac{emf^2}{R+r} ×t = [\frac{2×2}{5+1}]×[2×60]= 80 J$

b) Thermal energy

$U'=i^2Rt=[\frac{emf}{R+r}]^2 ×Rt=[\frac{2}{1+5}]^2×[5×2×60]= 67 J$

c) difference between a) and b)

The difference between U and U' which is 80-67=13J is the thermal energy that is generated in the battery due to internal resistance.