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: Phase angle of an RL circuit.
In a series resistor-inductor circuit, the current lags the voltage by $\phi$, where \[ \tan\phi = \frac{X_L}{R}. \]
Step 2: Inductive reactance.
The reactance is $X_L = \omega L = 2\pi f L$, with $f = 350\ \text{Hz}$ and $L = 0.1\ \text{H}$.
Step 3: Evaluate $X_L$.
\[ X_L = 2\pi(350)(0.1) = 70\pi. \] Taking $\pi \approx \dfrac{22}{7}$, \[ X_L = 70\times\frac{22}{7} = 220\ \Omega. \]
Step 4: Form the tangent.
With $R = 110\ \Omega$, \[ \tan\phi = \frac{220}{110} = 2. \]
Step 5: Solve for the phase angle.
\[ \phi = \tan^{-1}(2). \]
Step 6: Conclude.
The voltage maximum leads the current maximum by $\tan^{-1}(2)$.
\[ \boxed{\tan^{-1}(2)} \]