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  • A circuit element when connected to an a.c. supply of peak voltage <img alt="100 \mathrm{~V}" src="ht

A circuit element \mathrm{X} when connected to an a.c. supply of peak voltage 100 \mathrm{~V} gives a peak current of 5 \mathrm{~A} which is in phase with the voltage. A second element \mathrm{Y} when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by \frac{\pi}{2} .If \mathrm{X \: and \: Y} are connected in series to the same supply, what will be the rms value of the current in ampere?
 

Option: 1

\frac{10}{\sqrt{2}}


Option: 2

\frac{5}{\sqrt{2}}


Option: 3

5 \sqrt{2}


Option: 4

\frac{5}{2}


Answers (1)

best_answer

\mathrm{I_{1}=5A=\frac{100v}{R}}

\mathrm{R=20\Omega }

\mathrm{I_{1}}\rightarrow Peak current

Current and voltage are in same phase

\therefore Element X is resistor

Current lags behind the voltage by \mathrm{\pi/2}

Element Y\rightarrow Inductor

\mathrm{I_{2}=5A}

\mathrm{I_{2}}\rightarrowpeak current

\mathrm{X_{L}=20\Omega }

\mathrm{I}\rightarrow Peak current

\mathrm{I=\frac{V_0}{\sqrt{R^2+x_L^2}}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} }

\mathrm{I_{\text {rms }}=\frac{I}{\sqrt{2}}=\frac{5}{2} \mathrm{~A} }

Hence 4 is correct option.

Posted by

Divya Prakash Singh

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