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
Diffcult
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(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 Ohms law for resistance $R$, $V = iR$
$ \Rightarrow \,\,110 = 9 \times R$ $ \Rightarrow $ $R = 12.22\,\Omega $
$110 = \frac{{{V^2}}}{{110}}\,\, \Rightarrow \,V = 110\,V.$ 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 Ohms law for resistance $R$, $V = iR$
$ \Rightarrow \,\,110 = 9 \times R$ $ \Rightarrow $ $R = 12.22\,\Omega $
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