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

Explanation

The resistance of the heater is
`R = (V^(2))/(P)`
`= ( 100 xx 100 )/( 1000) = 10 Omega`
The power on which it operates is `62.5 W`. Therefore,
` V = sqrt( R xx P') = sqrt( 10 xx 62.5) = sqrt(625) = 25 V`
So the potential drop across ` AB` is `75 V`.
Therefore, the current in `AB` is
`I = (V) /(R) = ( 75)/(10) = 7.5 A`
This currents gets divided into two parts . Let `I_(1)` be the current that passes through the heater . Therefore,
` 25 = I_(1) xx 10 or I_(1) = 2.5 A`
Hence, the current through R is `5 A`.
Applying ohm's law across `R` , we get
` 25 = 5 xx R or R = 5 Omega`