Question

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50V.

Answer 1

we know that
heat = i^2r
=i×i×v/I
=iv
=q/t×v(I=q/t)

heat=96000/60×60×50
=1333.33joule

Answer 2

Answer:

Given charge Q = 96000 coulomb.

time = 1hr = 60 * 60 = 3600s.

Potential difference {V} = 50v

According to formula: H= VIT -(i)

Also I = Q/T

∴I  = 96000/3600 =80/3= Ampere

putting value of I = 80/3 Ampere in (i)

H=50 X 80/3 X 60 X 60 = 4.8 X 10²×³joule

Heat is generated that is 4.8 X 10²×³ joule