Question

A coil of inductance \(0.1\ \text{H}\) and resistance \(110\ \Omega\) is connected to a source of \(110\ \text{V}\) and \(350\ \text{Hz}\). The phase difference between the voltage maximum and the current maximum is

• A. \(\tan^{-1}(1.5)\)
• B. \(\tan^{-1}(0.5)\)
• C. \(\tan^{-1}(1.73)\)
• D. \(\tan^{-1}(2)\)

Hint:

In an \(RL\) circuit, current lags voltage by an angle \(\phi\), where \(\tan\phi=\dfrac{X_L}{R}\).

Answer 1

The Correct Option is D

Step 1: Use the formula for phase difference in an \(RL\) circuit.
For a series \(RL\) circuit, the phase difference between voltage and current is given by
\[ \tan\phi=\frac{X_L}{R} \]
where \(X_L\) is the inductive reactance and \(R\) is the resistance.

Step 2: Calculate inductive reactance.
Inductive reactance is
\[ X_L=\omega L \]
where
\[ \omega=2\pi f \]
Given,
\[ f=350\ \text{Hz} \] and
\[ L=0.1\ \text{H} \]
So,
\[ X_L=2\pi(350)(0.1) \]
\[ X_L=70\pi \]
Using \(\pi\approx\frac{22}{7}\),
\[ X_L=70\times\frac{22}{7} \]
\[ X_L=220\ \Omega \]

Step 3: Find the phase angle.
Given resistance,
\[ R=110\ \Omega \]
Therefore,
\[ \tan\phi=\frac{220}{110} \]
\[ \tan\phi=2 \]
Hence,
\[ \phi=\tan^{-1}(2) \]

Step 4: Final conclusion.
Thus, the phase difference between voltage maximum and current maximum is
\[ \boxed{\tan^{-1}(2)} \]