In an ac circuit, if voltage V = (a+jb) and current I = (c+jd), then power is given by
A
$ac+ad$
B
$ac+bd$
C
$bc-ad$
D
$bc+ad$

Question

In an ac circuit, if voltage V = (a+jb) and current I = (c+jd), then power is given by

Examoogle · Accepted Answer

$ac+bd$

Explanation: In an AC circuit, the complex power $S$ is defined as the product of the voltage phasor $V$ and the complex conjugate of the current phasor $I^*$. For $V = a+jb$ and $I = c+jd$, the conjugate is $I^* = c-jd$, so $S = (a+jb)(c-jd) = (ac+bd) + j(bc-ad)$.

Working Principle: The real part of the complex power, $P = Re(S)$, represents the active power dissipated in the circuit. Calculating the product $(a+jb)(c-jd)$ results in $ac - jad + jbc - j^2bd$. Since $j^2 = -1$, the term $-j^2bd$ becomes $+bd$, yielding a real component of $ac+bd$.

Key Formulas: $S = V \cdot I^*$; $P = Re(S) = ac+bd$; $Q = Im(S) = bc-ad$

Key Points: Complex power $S$ is measured in Volt-Amperes (VA).; The real part $P$ is the active power measured in Watts (W).; The imaginary part $Q$ is the reactive power measured in Volt-Amperes Reactive (VAR).; Taking the conjugate of the current is essential to maintain the correct phase relationship for reactive power.

Advantages: Allows unified analysis of resistive and reactive components.; Simplifies power flow calculations in steady-state AC systems.

Disadvantages: Requires understanding of complex numbers and phasor notation.

Applications: Power system load flow analysis.; Design of electrical machines and transmission lines.; AC motor power factor correction.

Additional Info: The active power is given by the real part of the product of voltage and the current conjugate.. Option A ($ac+ad$) is dimensionally incorrect for power components.. Option C ($bc-ad$) represents the reactive power $Q$ in the circuit.

Answer: B is correct — The real power in an AC circuit is defined by the real part of $V \cdot I^*$, which evaluates to $ac+bd$.

Exam Tip: Always remember the complex conjugate: when multiplying phasors for power, conjugate the current ($I^*$) to correctly account for the phase angle difference between voltage and current.

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