Question

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

Answer 1

Answer:

Given charge Q = 96000 coulomb.

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

Potential difference {V} = 50v

According to formula

${\boxed{\underline{\mathbf{\red{ H=VIT-(i)}}}}}$

Explanation:

Also I = $\frac{Q}{T}$

$\therefore I$ = $\frac{\cancel{96000}}{\cancel{3600}}$ = $\frac{80}{3}$Ampere

putting value of I = $\frac{80}{3}$Ampere in (i)

H=50 X $\frac{80}{3}$ X 60 X 60 = 4.8 X $10^6$joule

$\because$Heat is generated that is 4.8 X $10^6$joule

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