A heater is designed to operate with a power of `1000W` in a `100 V` line. It is connected in combination with a resistance of `10 Omega` and a resistance `R`, to a `100 V` mains as shown in figure. What will be the value of `R` so that the heater operates with a power of `62.5W`?

Options

A. ` 5 Omega`
B. `10 Omega`
C. `15 Omega`
D. `20 Omega`

Answer

C

Explanation

power of the heater is `P = 1000 W`. Potential difference is `V = 100 V`. Therefore , resistance is
`R_(1) = ( V^(2))/(P) = ( 100 xx 100)/(1000) = 10 Omega`
Now resistance of the circuit is
`R_(2) = 10 + ( 10 R) /( 10 + R) = ( 100 + 20 R) / ( 10 + R)`
Therefore , power is
`(V^(2))/( R_(2)) = 625 W`
or `R_(2) = (V^(2))/( 625) or ( 100 + 20 R )/( 10 + R) = (625)/( 100 xx 100) rArr R = 15 Omega`