Question

A heater coil is related 100 w ,200v . it is cut in two identical part and both parts are connected together in parallel to the same source of 200 volt calculate the energy library in 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

Answer 2

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