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

From `P=V^2/R` , Resistance of heater , `R=V^2/P=(100)^2/1000=10 Omega`
From `P=I^2R` Current required across heater for power of 62.5W
`i=sqrt(P/R) =sqrt(62.5/10)=2.5A`, Main current in the circuit
`I=100/(10+(10R)/(10+R))=(100(10+R))/(100+20R)=(10(10+R))/(10+2R)`
This current will distribute in inverse ratio of resistance between heater and R.
`i=(R/(10+R))I or 2.5 = (R/(10+R))[(10(10+R))(10+2R)]=(10R)/(10+2R)`
SOlving this equation we get `R=5 Omega`