Question

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`?

Answer 1

From `P=V^2/R`
Resistance of heater, `R=V^2/P=((100)^2)/1000=10Omega`
From `P=i^2 R`
Current required across heater of power of `62.5 W`,
`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=5Omega`