Question

An AC circuit has a resistance of 150 \(\Omega\), inductive reactance of 250 \(\Omega\), and capacitive reactance of 100 \(\Omega\). What is the phase difference between current and voltage?

• A. \( \tan^{-1}\left(\frac{250 - 100}{150}\right) \)
• B. \( \tan^{-1}\left(\frac{100 - 250}{150}\right) \)
• C. \( \tan^{-1}\left(\frac{150}{250 - 100}\right) \)
• D. \( \tan^{-1}\left(\frac{150}{250 + 100}\right) \)

Hint:

When calculating the phase difference in an RLC circuit, use the formula \(\tan(\phi) = \frac{X_L - X_C}{R}\) to find the angle of phase difference.

Answer 1

The Correct Option is A

In an AC circuit with a resistor, inductor, and capacitor, the phase difference \(\phi\) between the current and voltage is given by: \[ \tan(\phi) = \frac{X_L - X_C}{R} \] Where: - \(X_L\) is the inductive reactance (250 \(\Omega\)), - \(X_C\) is the capacitive reactance (100 \(\Omega\)), - \(R\) is the resistance (150 \(\Omega\)). Substitute the given values: \[ \tan(\phi) = \frac{250 - 100}{150} = \frac{150}{150} = 1 \] Thus, the phase difference is: \[ \phi = \tan^{-1}(1) = 45^\circ \] Therefore, the correct answer is option (A).