A heater is designed to operate with a power of 1000 walts in a 100 volt line. It is connected in a combinations with a resistance of 10 ohms and a resistance R to a 100 volts mains as shown in figure. What should be the value of R so that the heater operates with a power of 62.5 watts.
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Answer

A

Explanation

The resistance of the heater is
`R = V^2/P = (100 xx 100)/100 = 10Omega`
The power on which it operates is 62.5 W
`:. V = (sqrt(R xx P')) = (sqrt(10 xx 62.5)) = (sqrt(625)) = 25`
Since the voltage drop across the heater is 25V hence voltage drop across `10Omega` resistor is (100-25) = 75V.
`:. The current in AB = I = V/R = 75/10 = 7.5A`
This current divides into two parts. Let `I_1` be the current
that passes through the heater. Therefore,
`25 = I_1 xx 10 `
`I_1 = 2.5 A`
Thus current through R is 5A.

Applying Ohm's law across R, we get
`25 = 5 xx R`
`rArr R = 5Omega`.