Question

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

Answer 1

Answer:

we know that

H = i^2 × r × t. or. H = i×i×r×t

t = 1 hour

= 3600 seconds

q = 96000C

I = q/t

I = 96000/3600

I = 26.67A

V = ir

ir = 50

H = 26.67 × 50 × 3600

H = 4800600 J

Answer 2

Answer:

Given charge Q = 96000 coulomb.

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

Potential difference {V} = 50v

According to formula: H= VIT -(i)

Explanation:

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