Question 347: AC fundamentals

Question

An inductance coil has a resistance of $80 \Omega$. When on AC signal of frequency 480 Hz is applied to the coil, the voltage leads the current by $45^{\circ}$. The inductance of the coil in henry is $\left[\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}\right]$

Accepted Answer

To determine the inductance of the coil, we can use the concept of an R-L (resistance-inductance) circuit where the phase difference between voltage and current is given. The phase difference $\phi$ in an R-L circuit is given by:
$$ 	an \phi = \frac{X_L}{R} $$
where $X_L = \omega L$ is the inductive reactance, $R$ is the resistance, and $\omega = 2\pi f$ is the angular frequency. Given that the phase angle $\phi = 45^{\circ}$, we have:
$$ 	an 45^{\circ} = 1 $$
This implies:
$$ 1 = \frac{X_L}{R} $$
Substituting $X_L = \omega L$ and $R=80\,\Omega$, the equation becomes:
$$ 1 = \frac{\omega L}{80} $$
Thus:
$$ \omega L = 80 $$
Substitute $\omega = 2\pi f = 2\pi 	imes 480$:
$$ 2\pi 	imes 480 	imes L = 80 $$
Solving for $L$ gives:
$$ L = \frac{80}{2 \pi 	imes 480} = \frac{80}{960\pi} = \frac{1}{12\pi} $$
Therefore, the inductance of the coil is:
Option D: $\frac{1}{12 \pi}$ henry.

Answer

$\frac{1}{24 \pi}$

Answer

$\frac{\pi}{20}$

Answer

$\frac{\pi}{40}$

Answer

$\frac{1}{12 \pi}$

AC fundamentals

An inductance coil has a resistance of 80 \Omega. When on AC signal of frequency 480 Hz is applied to the coil, the voltage leads the current by 45^{\circ}. The inductance of the coil in henry is \left[\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}\right]

Choose the correct answer: