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AC i-V relationship for R, L, and C. Source v S (t) = Asin w t. Resistive Load. V R and i R are in phase. Phasor representation: v S (t) =Asin w t = Acos( w t-90 ° )= A -90 °= V S (j w ). I S (j w ) =(A / R) -90 °.

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ac i v relationship for r l and c
AC i-V relationship for R, L, and C

Source vS(t) =Asinwt

Resistive Load

VR and iR are in phase

Phasor representation: vS(t) =Asinwt = Acos(wt-90°)= A -90°=VS(jw)

IS(jw) =(A / R)-90°

Impendence: complex number of resistance Z=VS(jw)/ IS(jw)=R

Generalized Ohm’s law VS(jw) = Z IS(jw)

Everything we learnt before applies for phasors with generalized ohm’s law

capacitor load
Capacitor Load


VC(jw)= A -90°

Notice the impedance of a capacitance decreases with increasing frequency

inductive load
Inductive Load


Phasor: VL(jw)=A -90°

IL(jw)=(A/wL) -180°


Opposite to ZC, ZL increases with frequency

ac circuit analysis
AC circuit analysis
  • Effective impedance: example
  • Procedure to solve a problem
    • Identify the sinusoidal and note the excitation frequency.
    • Covert the source(s) to phasor form
    • Represent each circuit element by its impedance
    • Solve the resulting phasor circuit using previous learnt analysis tools
    • Convert the (phasor form) answer to its time domain equivalent.

Ex. 4.16, p180