Question 280: AC fundamentals

Question

When alternating current is passed through $\mathrm{L}-\mathrm{R}$ series circuit, the power factor is $\frac{\sqrt{3}}{2}$ and $\mathrm{R}=50 \Omega$, then the value of L is
$$ \left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \sin \frac{\pi}{6}=\frac{1}{2}, 	an \frac{\pi}{6}=\frac{1}{\sqrt{3}}ight] $$

Accepted Answer

$ \cos \phi = \frac{R}{Z} $
where $Z = \sqrt{R^2 + X_L^2}$, and $X_L = \omega L$.
Alternatively,
$ 	an \phi = \frac{X_L}{R}. $
Step 2. Find $\phi$.
Given:
$ \cos \phi = \frac{\sqrt{3}}{2}. $
So,
$ \phi = \frac{\pi}{6}. $
Then
$ 	an \phi = \frac{1}{\sqrt{3}}. $
Step 3. Use $	an \phi = X_L / R$.
$ \frac{X_L}{R} = \frac{1}{\sqrt{3}} \implies X_L = \frac{R}{\sqrt{3}}. $
Given $R = 50$:
$ X_L = \frac{50}{\sqrt{3}}. $
Step 4. Relating $X_L$ to $L$.
$ X_L = \omega L. $
Usually, unless otherwise stated, $\omega = 2\pi f$. For standard power-frequency problems,
it is implicitly assumed $f = 50 \,	ext{Hz}$, thus $\omega = 2\pi f = 100\pi \,	ext{rad/s}$.
So
$ L = \frac{X_L}{\omega} = \frac{50/\sqrt{3}}{100\pi}. $
Step 5. Simplify.
$ L = \frac{50}{100 \sqrt{3} \, \pi} = \frac{1}{2\sqrt{3}\pi}. $
Final Answer:
$ \boxed{	frac{1}{2\sqrt{3}\pi}} $
Correct option: C.

Answer

$\frac{1}{2} \pi$

Answer

$\frac{\sqrt{3}}{2} \pi$

Answer

$\frac{1}{2 \sqrt{3} \pi}$

Answer

$\frac{1}{\sqrt{3} \pi}$

AC fundamentals

When alternating current is passed through \mathrm{L}-\mathrm{R} series circuit, the power factor is \frac{\sqrt{3}}{2} and \mathrm{R}=50 \Omega, then the value of L is $ \left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \sin \frac{\pi}{6}=\frac{1}{2}, \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]

Choose the correct answer: