Capacitor A circuit element that stores electric energy and electric charges A capacitor always consists of two separated metals, one stores +q, and the other stores –q. A common capacitor is made of two parallel metal plates. Capacitance is defined as: C=q/V (F); Farad=Colomb/volt Once the geometry of a capacitor is determined, the capacitance (C) is fixed (constant) and is independent of voltage V. If the voltage is increased, the charge will increase to keep q/V constant Application: sensor (touch screen, key board), flasher, defibrillator, rectifier, random access memory RAM, etc.

Capacitor: cont. Because of insulating dielectric materials between the plates, i=0 in DC circuit, i.e. the braches with Cs can be replaced with open circuit. However, there are charges on the plates, and thus voltage across the capacitor according to q=Cv. i-v relationship: i = dq/dt = C dv/dt Solving differential equation needs an initial condition Energy stored in a capacitor: W C =1/2 Cv C (t) 2

Capacitors in V=V 1 =V 2 =V 3 q=q 1 +q 2 +q 3 parallel series V=V 1 +V 2 +V 3 q=q 1 =q 2 =q 3

Inductor i-v relationship: v L (t)= Ldi L /dt L: inductance, henry (H) Energy stored in inductors W L = ½ Li L 2 (t) In DC circuit, can be replaced with short circuit

Sinusoidal waves Why sinusoids: fundamental waves, ex. A square can be constructed using sinusoids with different frequencies (Fourier transform). x(t)=Acos(  t+  ) f=1/T cycles/s, 1/s, or Hz  =2  f rad/s  2  t /  rad =360  t /  deg.

Average and RMS quantities in AC Circuit It is convenient to use root-mean-square or rms quantities to indicate relative strength of ac signals rather than the magnitude of the ac signal.

Complex number review Euler’s indentity a b

Phasor How can an ac quantity be represented by a complex number? Acos(  t+  )=Re(Ae j(  t+  ) )=Re(Ae j  t e j  ) Since Re and e j  t always exist, for simplicity Acos(  t+  )  Ae j   Phasor representation Any sinusoidal signal may be mathematically represented in one of two ways: a time-domain form v(t) = Acos(  t+  ) and a frequency-domain (or phasor) form V(j  ) = Ae j   In text book, bold uppercase quantity indicate phasor voltage or currents Note the specific frequency  of the sinusoidal signal, since this is not explicit apparent in the phasor expression

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 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

R1=100 , R 2 =75 , C= 1  F, L=0.5 H, v S (t)=15cos(1500t) V