A heater is designed to operate with a power of 1000 watt 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

Let R. be the resistance of the heater of coil.
`R=(V^(2))/(P)=((100)^(2))/(1000)=10Omega`
If the heater has to operate with a power `P=62.5W`, the voltage V. across its coil should be
`V.=(P.R.)^(1//2)=(6.25xx10)^(1//2)=25V`
Thus, out of 100 V, a voltage drop of 25 V occurs across the heater and the rest `100-25=75V` occurs across the `10Omega` resistor. Therefore, current in the circuit is
`I=(75)/(10)=7.5A`
Now, current through the heater `=(V.)/(R.)=(25)/(10)=2.5A`
Therefore, current through resistor
`R=7.5-2.5=5.0A`
`"Hence, "R=(V.)/(5.0A)=(25V)/(5.0A)=5Omega`