For an RL circuit, the phase angle is always

Question

Examoogle · Accepted Answer

Negative

Explanation: In an RL circuit, the current lags behind the voltage by a phase angle $\phi$. By convention, when the current lags the voltage, the phase angle is considered positive, but the impedance phase angle is defined as $	an^{-1}(\frac{X_L}{R})$, which is positive for inductors, leading to a negative phase displacement for the current.

Working Principle: The inductor opposes changes in current, causing the current vector to rotate clockwise relative to the voltage vector in a phasor diagram. Since the voltage $V = I(R + jX_L)$, the phase angle of the impedance is $	heta = \arctan(\frac{X_L}{R})$, which is always positive. The phase angle of current relative to voltage is defined as $-	heta$.

Key Formulas: $\phi = \arctan(\frac{\omega L}{R})$ — Phase angle of the impedance; $\vec{Z} = R + jX_L$ — Complex impedance of an RL circuit

Key Points: Inductive circuits always have a lagging power factor.; The impedance angle $\phi$ is positive because $X_L > 0$.; Current lags voltage by angle $\phi$, often represented as $-j\phi$ in phasor notation.; In an ideal RL circuit, $0

Advantages: Used in filtering applications; Essential for inductive load modeling

Disadvantages: Causes lagging reactive power consumption; Reduces the overall power factor of the system

Applications: Induction motors; Transformers; Choke coils

Additional Info: The question implies the phase shift of current with respect to voltage. Since current lags, the displacement is negative.. If the question referred to the impedance phase angle, the answer would be positive. Always context-check whether referencing Impedance or Current-relative-to-Voltage.

Answer: B is correct — In an RL circuit, the current lags the voltage, resulting in a negative phase shift.

Exam Tip: Always remember: 'ELI the ICE man'. In 'ELI', E (voltage) leads I (current) in an L (inductor) circuit, meaning I lags E by a negative angle.

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