Question 9: AC fundamentals

Question

A pure resistive circuit element $X$, when connected to an AC supply of peak voltage 200 V , gives a peak current of 5 A . A second circuit element $Y$, when connected to the same AC supply also gives the same value of peak current but the current lags behind by $90^{\circ}$. If the series combination of $X$ and $Y$ is connected to the same supply, what will be the rms value of current?

Accepted Answer

Clearly first element is resistance and second element is inductor
$$ \begin{aligned} R & =X_L=\frac{200}{5}=40 \Omega \ 	herefore Z & =\sqrt{R_f^2+X_L^2}=40 \sqrt{2} \Omega \ I_{\mathrm{rms}} & =\frac{V_{\mathrm{rms}}}{Z}=\frac{200 / \sqrt{2}}{40 \sqrt{2}}=\frac{5}{2} \mathrm{~A} \end{aligned} $$

Answer

$\frac{10}{\sqrt{2}} \mathrm{~A}$

Answer

$\frac{5}{\sqrt{2}} \mathrm{~A}$

Answer

$\frac{5}{2} \mathrm{~A}$

Answer

5 A

AC fundamentals

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