Question 165: AC fundamentals

Question

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 $$\mathrm{Y}$$ are connected in series to the same supply, what will be the rms value of the current in ampere?

Accepted Answer

Element X should be resistive with, $R=\frac{100}{5}=20 \Omega$
Element Y should be inductive with, $$
X_{L}=\frac{100}{5}=20 \Omega
$$
When X and Y are connector in series,
$$
\begin{aligned}
&Z=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \Omega \\
&I=\frac{100}{Z}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} \\
&i_{\mathrm{rms}}=\frac{1}{\sqrt{2}} I \\
&=\frac{5}{2}
\end{aligned}
$$

Answer

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

Answer

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

Answer

$$5 \sqrt{2}$$

Answer

$$\frac{5}{2}$$

AC fundamentals

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