Question

A coil of inductive reactance $\frac{1}{\sqrt{3}} \Omega$ and a resistance 1 $\Omega$ are connected in series to a 200 V, 50 Hz ac source. The time lag between voltage and current is

• A. 1200 s
• B. 1 s
• C. $\frac{\pi}{600}$ s
• D. $\frac{\pi}{1800}$ s

Hint:

In an LR circuit, the phase difference between voltage and current depends on the ratio of inductive reactance to resistance, and the time lag is the phase angle divided by the angular frequency.

Answer 1

The Correct Option is C

Let’s break this down step by step to calculate the time lag between voltage and current in the LR circuit and determine why option (3) is the correct answer.
Step 1: Understand the concept of time lag in an LR circuit
In an LR circuit with an AC source, the voltage leads the current by a phase angle $\phi$, where:
\[ \tan \phi = \frac{X_L}{R} \]

• $X_L$ is the inductive reactance,
• $R$ is the resistance.

The time lag $\Delta t$ between voltage and current is related to the phase angle by:
\[ \Delta t = \frac{\phi}{\omega} \]
where $\omega = 2\pi f$ is the angular frequency, and $f$ is the frequency of the AC source.
Step 2: Identify the given values and calculate the phase angle

• Inductive reactance, $X_L = \frac{1}{\sqrt{3}} \, \Omega$
• Resistance, $R = 1 \, \Omega$
• Frequency, $f = 50 \, \text{Hz}$

\[ \tan \phi = \frac{X_L}{R} = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}} \]
\[ \phi = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = 30^\circ = \frac{\pi}{6} \, \text{radians} \]
Step 3: Calculate the angular frequency and time lag
\[ \omega = 2\pi f = 2\pi \times 50 = 100\pi \, \text{rad/s} \]
\[ \Delta t = \frac{\phi}{\omega} = \frac{\frac{\pi}{6}}{100\pi} = \frac{\pi}{6 \times 100\pi} = \frac{\pi}{600} \, \text{s} \]
Step 4: Confirm the correct answer
The calculated time lag is $\frac{\pi}{600} \, \text{s}$, which matches option (3).
Thus, the correct answer is (3) $\frac{\pi}{600}$ s.