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AC i-V relationship for R, L, and C

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

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  1. 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

  2. Capacitor Load ICE VC(jw)= A -90° Notice the impedance of a capacitance decreases with increasing frequency

  3. Inductive Load ELI Phasor: VL(jw)=A -90° IL(jw)=(A/wL) -180° ZL=jwL Opposite to ZC, ZL increases with frequency

  4. 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

  5. Ex. 4.21 P188 R1=100 W, R2=75 W, C= 1mF, L=0.5 H, vS(t)=15cos(1500t) V. Determine i1(t) and i2(t). Step 1: vS(t)=15cos(1500t), w=1500 rad/s. Step 2: VS(jw)=15 0 Step 3: ZR1=R1, ZR2=R2, ZC=1/jwC, ZL=jwL Step 4: mesh equation

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