AC i-V relationship for R, L, and C Resistive Load Source v S (t)  Asin  t V R and i R are in phase Phasor representation: v S (t) =Asin  t = Acos(  t-90°)= A  -90°=V S (j  ) I S (j w ) =(A / R)  -90° Impendence: complex number of resistance Z=V S (j  )/ I S (j  )=R Generalized Ohm’s law V S (j  ) = Z I S (j  ) Everything we learnt before applies for phasors with generalized ohm’s law

Capacitor Load ICE V C (j  )= A  -90° Notice the impedance of a capacitance decreases with increasing frequency

Inductive Load Phasor: V L (j  -90° I L (j  )=( A/  L)  -180° Z L =j  L ELI Opposite to Z C, Z L increases with frequency

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

Ex P188 R 1 =100  R 2 =75  C= 1  F, L=0.5 H, v S (t)=15cos(1500t) V. Determine i 1 (t) and i 2 (t). Step 1: v S (t)=15cos(1500t),  =1500 rad/s. Step 2: V S (j  )=15  0 Step 3: Z R1 =R 1, Z R2 =R 2, Z C =1/j  C, Z L =j  L Step 4: mesh equation