Question 294: AC fundamentals

Question

A coil of resistance $450 \Omega$ and self-inductance 1.5 henry is connected to an a.c. source of frequency $\frac{150}{\pi} \mathrm{~Hz}$. The phase difference between voltage and current is

Accepted Answer

Given:
Resistance, $R = 450\,\Omega$
Self-inductance, $L = 1.5\,	ext{H}$
Frequency, $f = \dfrac{150}{\pi}~	ext{Hz}$
Step 1: Calculate angular frequency $\omega$:
$ \omega = 2\pi f = 2\pi 	imes \dfrac{150}{\pi} = 2 	imes 150 = 300~	ext{rad/s} $
Step 2: Inductive reactance $X_L$:
$ X_L = \omega L = 300 	imes 1.5 = 450~\Omega $
Step 3: Phase difference formula:
The phase difference $\phi$ between voltage and current in a series $LR$ circuit is given by:
$ 	an\phi = \frac{X_L}{R} $
Substituting the values:
$ 	an\phi = \frac{450}{450} = 1 $
Step 4: Hence,
$ \phi = 	an^{-1}(1) $
Correct Option: Option B
$ \boxed{	an^{-1}(1)} $

Answer

$	an ^{-1}(0.5)$

Answer

$	an ^{-1}(1)$

Answer

$	an ^{-1}(1.5)$

Answer

$	an ^{-1}(2.0)$

AC fundamentals

A coil of resistance 450 \Omega and self-inductance 1.5 henry is connected to an a.c. source of frequency \frac{150}{\pi} \mathrm{~Hz}. The phase difference between voltage and current is

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