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

A

ac+adac+adac+ad

B

ac+bdac+bdac+bd

C

bcтИТadbc-adbcтИТad

D

bc+adbc+adbc+ad

Correct Answer

тЪЩя╕П TE тАв Technical Concept & PrincipleElectricalBasic Electrical
Option B

ac+bdac+bdac+bd

Quick Summary:

In an AC circuit, the complex power SSS is defined as the product of the voltage phasor VVV and the complex conjugate of the current phasor IтИЧI^*IтИЧ. For V=a+jbV = a+jbV=a+jb and I=c+jdI = c+jdI=c+jd, the conjugate is IтИЧ=cтИТjdI^* = c-jdIтИЧ=cтИТjd, so S=(a+jb)(cтИТjd)=(ac+bd)+j(bcтИТad)S = (a+jb)(c-jd) = (ac+bd) + j(bc-ad)S=(a+jb)(cтИТjd)=(ac+bd)+j(bcтИТad).

тЪЩя╕ПTETechnical SolutionConcept & Principle
ЁЯТб Explanation

In an AC circuit, the complex power SSS is defined as the product of the voltage phasor VVV and the complex conjugate of the current phasor IтИЧI^*IтИЧ. For V=a+jbV = a+jbV=a+jb and I=c+jdI = c+jdI=c+jd, the conjugate is IтИЧ=cтИТjdI^* = c-jdIтИЧ=cтИТjd, so S=(a+jb)(cтИТjd)=(ac+bd)+j(bcтИТad)S = (a+jb)(c-jd) = (ac+bd) + j(bc-ad)S=(a+jb)(cтИТjd)=(ac+bd)+j(bcтИТad).

ЁЯФв Key Formulas

S=VтЛЕIтИЧS = V \cdot I^*S=VтЛЕIтИЧ

P=Re(S)=ac+bdP = Re(S) = ac+bdP=Re(S)=ac+bd

Q=Im(S)=bcтИТadQ = Im(S) = bc-adQ=Im(S)=bcтИТad

тЪЩя╕П Working Principle

The real part of the complex power, P=Re(S)P = Re(S)P=Re(S), represents the active power dissipated in the circuit. Calculating the product (a+jb)(cтИТjd)(a+jb)(c-jd)(a+jb)(cтИТjd) results in acтИТjad+jbcтИТj2bdac - jad + jbc - j^2bdacтИТjad+jbcтИТj2bd. Since j2=тИТ1j┬▓ = -1j2=тИТ1, the term тИТj2bd-j^2bdтИТj2bd becomes +bd+bd+bd, yielding a real component of ac+bdac+bdac+bd.

ЁЯУМ Key Points
  • тЦ╕

    Complex power SSS is measured in Volt-Amperes (VA).

  • тЦ╕

    The real part PPP is the active power measured in Watts (W).

  • тЦ╕

    The imaginary part QQQ is the reactive power measured in Volt-Amperes Reactive (VAR).

  • тЦ╕

    Taking the conjugate of the current is essential to maintain the correct phase relationship for reactive power.

тЬЕ Advantages
  • тЦ╕

    Allows unified analysis of resistive and reactive components.

  • тЦ╕

    Simplifies power flow calculations in steady-state AC systems.

тЭМ Disadvantages / Limitations
  • тЦ╕

    Requires understanding of complex numbers and phasor notation.

  • тЦ╕

    Not applicable to non-sinusoidal waveforms without Fourier transformation.

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Power system load flow analysis.

  • тЦ╕

    Design of electrical machines and transmission lines.

  • тЦ╕

    AC motor power factor correction.

ЁЯУД Additional Information
  • тЦ╕

    The active power is given by the real part of the product of voltage and the current conjugate.

  • тЦ╕

    Option A (ac+adac+adac+ad) is dimensionally incorrect for power components.

  • тЦ╕

    Option C (bcтИТadbc-adbcтИТad) represents the reactive power QQQ in the circuit.

ЁЯУК Diagram / Illustration
Complex Power CalculationS = V ├Ч I*V = a+jb, I = c+jdActive Power P = ac+bd
тЬЕ

B is correct тАФ The real power in an AC circuit is defined by the real part of VтЛЕIтИЧV \cdot I^*VтЛЕIтИЧ, which evaluates to ac+bdac+bdac+bd.

Core Concepts Used
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Phasor Algebra Complex Power Active and Reactive Power
ЁЯТб EXAM TIP

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

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