Question 394: 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: