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  An electric lamp of 100\: \Omega, a toaster of resistance 50\: \Omega and a water filter of resistance 500\: \Omegaare connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

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Given : R_1=100\Omega ,R_2=50\Omega ,R_3=500\Omega

    R=Equivalent  resistance

    \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} 

\Rightarrow \frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}

\Rightarrow \frac{1}{R}=\frac{5+10+1}{500}=\frac{16}{500}

\Rightarrow R=\frac{500}{16}=31.25\Omega

By Ohm's law,

                   I=\frac{V}{R}=\frac{220}{31.25}=7.04A

Hence, the resistance of electric iron is 31.25 and current through it is 7.04A.

 

 

 

 

 

Posted by

seema garhwal

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