Question

Two lamps, one rated 100 W at 220 V and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V? ​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

Two lamps, one rated 100 watt at 220 V and the other 60 watt at 220 V are connected in parallel to electric mains supply.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The current is drawn from the line if the supply voltage is 220 V.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Resistance of first lamp = $\sf{R_{1}}$

Resistance of second lamp = $\sf{R_{2}}$

We know that formula of the resistance :

$\boxed{\bf{R=\frac{V^{2} }{P} }}}$

A/q

$\longrightarrow\tt{R_{1}=\dfrac{220\times 220}{100} }\\\\\\\longrightarrow\tt{R_{1}=\cancel{\dfrac{48400}{100} }}\\\\\\\longrightarrow\tt{\pink{R_{1}=484\:\Omega}}$

&

$\longrightarrow\tt{R_{2}=\dfrac{220\times 220}{60} }\\\\\\\longrightarrow\tt{R_{2}=\cancel{\dfrac{48400}{60} }}\\\\\\\longrightarrow\tt{\pink{R_{2}=\dfrac{2420}{3} \:\Omega}}$

$\underline{\underline{\bf{When\:R_{1}\:and\:R_{2}\:are\:conected\:in\:parallel\::}}}$

$\implies\tt{\dfrac{1}{R} =\dfrac{1}{R_{1}} +\dfrac{1}{R_{2}}\:\:\:\bigg[R\:is\:equivalent\:resistance\bigg]}\\\\\\\implies\tt{R=\dfrac{R_1R_2}{R_1+R_2} }\\\\\\\implies\tt{R=\dfrac{484\times \dfrac{2420}{3} }{484+\dfrac{2420}{3} } }\\\\\\\implies\tt{R=\dfrac{\cancel{484}\times \dfrac{2420}{3} }{\cancel{484}\bigg(1+\dfrac{5}{3} \bigg)}}\\\\\\\implies\tt{R=\dfrac{\dfrac{2420}{3} }{\dfrac{3+5}{3} } }\\\\\\\implies\tt{R=\dfrac{\dfrac{2420}{3} }{\dfrac{8}{3} }}$

$\implies\tt{R=\dfrac{2420}{\cancel{3}} \times \dfrac{\cancel{3}}{8} }\\\\\\\implies\tt{R=\cancel{\dfrac{2420}{8} }}\\\\\\\implies\tt{\pink{R=\dfrac{605}{2} \:\Omega}}$

Now;

We know that formula of the current (i) :

$\boxed{\bf{i=\frac{V }{R} }}}$

$\longrightarrow\sf{i=\dfrac{220}{\dfrac{605}{2} }} \\\\\\\longrightarrow\sf{i=\dfrac{\cancel{220}\times 2}{\cancel{605}} }\\\\\\\longrightarrow\sf{i=\dfrac{20\times 2}{55} }\\\\\\\longrightarrow\sf{i=\cancel{\dfrac{40}{55} }}\\\\\\\longrightarrow\sf{\pink{i=0.727\:Ampere}}$

Thus;

$\underbrace{\sf{The\:\boxed{\bf{0.727\:ampere\:(A)}}}\:current\:is\:drawn\:from\:the\:line\:if\:the\:supply\:voltage\:220\:volt}}}}}$