Question 381: AC fundamentals

Question

A coil having an inductance of $\frac{1}{\pi} \mathrm{H}$ is connected in series with a resistance of $300 \Omega$. If A.C. Source $(20 \mathrm{~V}-200 \mathrm{~Hz})$ is connected across the combination, the phase angle between voltage and current is

Accepted Answer

The phase angle $ \phi $ in an R-L circuit (a circuit with resistance and inductance) is determined by the ratio of the inductive reactance ($ X_L $) to the resistance ($ R $). The inductive reactance is given by the formula:
$
X_L = 2 \pi f L
$
where:
$ f $ is the frequency of the alternating current (in hertz).
$ L $ is the inductance of the coil (in henrys).
Given that the inductance $ L = \frac{1}{\pi} \, \mathrm{H} $ and the frequency $ f = 200 \, \mathrm{Hz} $, we can calculate $ X_L $ as follows:
$
X_L = 2 \pi \cdot 200 \cdot \frac{1}{\pi} = 2 \cdot 200 = 400 \, \Omega
$
The phase angle $ \phi $ is the arctangent of the ratio of inductive reactance to resistance:
$
\phi = 	an^{-1} \left( \frac{X_L}{R} ight)
$
Since the resistance $ R = 300 \, \Omega $, we can insert the values into the formula to find the phase angle:
$
\phi = 	an^{-1} \left( \frac{400 \, \Omega}{300 \, \Omega} ight) = 	an^{-1}\left(\frac{4}{3}ight)
$
Therefore, the correct answer is: Option B $ 	an ^{-1}\left(\frac{4}{3}ight) $.

Answer

$	an ^{-1}\left(\frac{3}{4}ight)$

Answer

$	an ^{-1}\left(\frac{4}{3}ight)$

Answer

$	an ^{-1}\left(\frac{5}{4}ight)$

Answer

$	an ^{-1}\left(\frac{4}{5}ight)$

AC fundamentals

A coil having an inductance of \frac{1}{\pi} \mathrm{H} is connected in series with a resistance of 300 \Omega. If A.C. Source (20 \mathrm{~V}-200 \mathrm{~Hz}) is connected across the combination, the phase angle between voltage and current is

Choose the correct answer: