Question

A heater coil is rated 100w,200v. It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200v. Calulate the energy liberated per second in the new combination.

Answer 1

Resistance of the heater coil is found using,

P = V²/ R

or

R = V²/ P = 220² /100

thus, resistance will be

R = 484 Ω

Now, when we cut it into half, the resistance of each part is R/2 = 242 Ω, the equivalent resistance across the ends is

= (R/2) / 2 = 121 Ω

So, now the power dissipated as heat is, P = 220² / 121

thus, power (energy liberated per second) will be

P = 400 W