Question

In sinusoidal steady-state analysis, which of the following best describes the use of phasors?

• A. Phasors are used to simplify time-domain analysis of RLC circuits by representing sinusoidal signals as complex numbers
• B. Phasors are only used for DC analysis
• C. Phasors convert differential equations into algebraic equations in the time domain
• D. Phasors are used to compute the time derivative of voltage or current

Hint:

Phasors transform complex calculus operations in the time domain (\(\frac{d}{dt} \rightarrow j\omega\)) into simple algebraic multiplications in the frequency domain using complex values.

Answer 1

The Correct Option is A

Understanding the Concept: A phasor is a complex number that represents the amplitude and initial phase angle of a sinusoidal steady-state signal. In time-domain analysis, RLC circuits are governed by linear integro-differential equations, which can be tedious and complex to solve directly. By applying Euler's relation, a sinusoidal signal \(v(t) = V_m \cos(\omega t + \phi)\) can be mapped into the frequency domain as a fixed complex number: \[ V = V_m e^{j\phi} = V_m \angle\phi \] This transformation eliminates the time dependence factor (\(t\)) from the active calculation process, effectively converting linear differential equations into manageable, straightforward complex algebraic expressions.

Step 1: Evaluating Option (A) in detail.
Option (A) states that phasors simplify the time-domain analysis of complex reactive networks by representing sinusoidal time-varying components as static complex vector expressions. This is the exact definition and primary operational purpose of utilizing phasor notation in electrical engineering. Therefore, Option (A) is completely accurate.

Step 2: Analyzing why the alternative options are incorrect.

• Option (B) is false because DC analysis involves constant values where frequency \(\omega = 0\). Phasors are explicitly tailored for alternating current (AC) sinusoidal steady-state operations.
• Option (C) is false because phasors convert time-domain differential equations into algebraic expressions in the frequency domain, not within the time domain itself.
• Option (D) is false because while taking a time derivative transforms a phasor by multiplying it by \(j\omega\), the fundamental reason we define and use phasors is to map the entire signal framework into complex numeric representations to bypass solving calculus equations directly. Thus, Option (A) is uniquely correct.