The resistance of a heater coil is `110 ohm`. A resistance `R` is connected in parallel with it and the combination is joined in series with a resistance of `11 ohm` to a 220 volt main line. The heater operates with a power of 110 watt. The value of `R` in ohm is

Options

A. 12.22
B. 24.22
C. Negative
D. That the given values are not correct

Answer

A

Explanation

(a) Power consumed by heater is `110 W` so by using
`P = (V^(2))/(R )`
` 110 = (V^(2))/(110) implies V = 110 V`. Also from figure `i_(1) = (110)/(110) =1 A` ltbRgt and `i= (110)/(11) = 10 A`. So `i_(2) = 10 - 1 = 9 A`
Appying Ohms laws for resistance `R, V = iR`
`implies 110 = 9 xx R implies T = 12.22 Omega`