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. `10 Omega`
B. `62.5 Omega`
C. `1/5 Omega`
D. `5 Omega`

Answer

D

Explanation

`R_(H) = V^(2)/P =(100)^(2)/1000 = 10 Omega`
`R. = 10 + (10 xx R)/(10 + R)`
`I_(H) = I(R/(10 + R)) =100/((100 + 20R)/(10 + R)) xx R/(10 + R) = (5R)/(5 + R)`
`therefore P = I_(H)^(2)R_(H) = (25 R^(2))/(5 +R)^(2) xx 10 = 62.5 W`
`therefore R = 5 Omega`