Question 418: AC fundamentals

Question

An AC circuit contains resistance of $12 \Omega$ and inductive reactance $5 \Omega$. The phase angle between current and potential difference will be

Accepted Answer

Given, $R=12 \Omega$ and $X_L=5 \Omega$
$	herefore$ Impedance, $Z=\sqrt{(12)^2+(5)^2}=13 \Omega$
The impedance triangle is as shown below
From this triangle,
$$\begin{aligned}
& \cos 	heta=\frac{R}{Z} \
& \Rightarrow \quad 	heta=\cos ^{-1}\left(\frac{R}{Z}ight)=\cos ^{-1}\left(\frac{12}{13}ight)
\end{aligned}$$

Answer

$\sin ^{-1}\left(\frac{12}{13}\right)$

Answer

$\cos ^{-1}\left(\frac{5}{12}\right)$

Answer

$\sin ^{-1}\left(\frac{5}{12}\right)$

Answer

$\cos ^{-1}\left(\frac{12}{13}\right)$

AC fundamentals

An AC circuit contains resistance of 12 \Omega and inductive reactance 5 \Omega. The phase angle between current and potential difference will be

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