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

Given,

t = 1 hour = 3600 s
V = 50 V
Q = 96000 C

I = $\dfrac{\text{Q}}{\text{t}} = \dfrac{96000}{3600} = \dfrac{80}{3}$A

According to Joule's law :

Heat (H) = Voltage (V) x current (I) x time (t)

Substituting we get,

H = 50 x $\dfrac{80}{3}$ x 3600 = 4.8 x 10⁶ J

Therefore, amount of heat generated = 4.8 x 10⁶ J