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

Answer

A

Explanation

Resistance of heater `R_(h ) = ((100)^(2))/(1000) = 10 Omega`
`P_(h) = i_(h)^(2) R_(h) implies 62.5 = i_(h)^(2)xx10`
`i_(h)= 2.5A`

`5R = 25`
`R = 5 Omega`.