Question

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

Answer 1

Answer: (A)

Power consumed by heater is $110 \,W$, so by using $P=\frac{V^{2}}{R}$

$110=\frac{V^{2}}{110} $
$\Rightarrow V=110\, V$.
Also from figure $i_{1}=\frac{110}{110}=1 A$
and $i=\frac{110}{11}=10 A .$
So $i_{2}=10-1=9 A$
Applying Ohm's law for resistance
$R, V=i R$
$\Rightarrow 110=9 \times R $
$\Rightarrow R=12.22 \,\Omega$