Q. 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
A) $12.22$
B) $24.42$
C) Negative
D) That the given values are not correct
Explanation
(a) Power consumed by heater is $110\ W,$ so by using $P = \frac{V^{2}}{R}$
$110 = \frac{V^{2}}{110} \Rightarrow V = 110V.\ $Also from figure $i_{1} = \frac{110}{110} = 1A$
and $\ i = \frac{110}{11} = 10\ A.$ So$\ i_{2} = 10 - 1 = 9\ A$
Applying $Ohms$ law for resistance $R,\ V = iR$
$\Rightarrow 110 = 9 \times R \Rightarrow R = 12.22\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.