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 I10 watt. The value of R in ohm is

# 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 I10 watt. The value of R in ohm is

1. A

12.22

2. B

24.42

3. C

Negative

4. D

That the given values are not correct

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### Solution:

Power consumed by heater is 110 W, so by using

$\mathrm{P}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$

$110\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{V}}^{2}}{110}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\mathrm{V}=110\mathrm{V}.$ Also from figure, ${\mathrm{i}}_{1}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{110}{110}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}1\mathrm{A}$ and $i\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{110}{11}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}10A$.  $\mathrm{So}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{i}}_{2}\text{\hspace{0.17em}}=10-1=9\mathrm{A}$

Applying Ohm's law for resistance R, V = iR

$⇒\text{\hspace{0.17em}\hspace{0.17em}}110=9\text{\hspace{0.17em}\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{R}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}=12.22\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$

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