Question

HW5 Problem 1 Creating Symbolic Expressions My Solutions > Problemi In the circuit below, the voltage source is given by v(t) = 12cos(400t - 30°). R1 = R2 = R3 = 51 and L1 = L2 = 20mH and C = mF. Please answer the following questions. w LI NL2 m i4 w un i3 R3 a) Transform the circuit into phasor domain b) Write out KCL for node N1 and N2 in the phasor domain c) Use KVL to express the currents in terms of node voltages in the phasor domain. The node voltages V1 and V2 are the voltage drops from N1 and N2 to the ground. d) Write the equations you derived in part (b) and (c) in a matrix form. e) Solve the systems of linear equations you derived in part (d) with any MATLAB, and then find syms amp_ic ph_ic f_ic Script Save C Reset MATLAB Documentation 1 syms amp_ic ph_ic f_ic 2 amp_ic = []; % Please input the amplitude of ic(t) 3 ph_ic = []; Please input the phase angle of ic(t) (In degree.) 4 f_ic = []; % Please input the angular frequency of ic(t) (In rad/s.)

Answer 1

Let w=frequency in rad / sec of reference phasor

TIME DOMAIN to PHASOR - DOMAIN

Resistance (R) ----------------------> R

Inductive Impedance (L) ----------------------> jwL

Capacitive Impedance (C) --------------------> 1/(jwC)

Basics

Kirchhoff's Current Law: For a lumped network, algebraic sum of currents entering (or leaving) a node is equal to zero for every instant of time.

Kirchhoff's Voltage Law: For a lumped network, algebraic sum of voltages in any close path is equal to zero for every instant of time.

Lumped network is a network in which the network components i.e., R,L and C can be separated.

. j = 1 290° complex operatori that shifts waveform s by 90° anticlockwise in phasor domain, II Vz - u - z - Applying KUL na KUL V, - Vz_ I → z Applykch, I, + 12 = I3

Now, Let us solve the problem.

Vet) = 12 cos (400t -- 30° ). W=400' rad /second, Let us take cos (40ot) as reference phasor, So, v= 12 L-30° Xi- j w Li = jk u U0 x 20x10 -3 = jer XL2 = jwLzz jer Xc. Iliwe - jx yoox 10 x 10-3 J 16 -) Xc=-jur (a) Redrawing circuit in phasor comcan. 11 BA T R1252 Iun 12 yea Ni - Ty mom XL2=jsrl V T o X4=jB2 - I SR3-512 ..to V 12-30°

b) Applying KCL in nude Al. Il + 14+ Ic = o T O KCL in nocle Me I1 + Iu= Izt is 1. Again, I 2 = Iz (6) Apply KUL in node a No, t his n o From equation u - V, V +.NI-V2 TV-o zon Ri . xu. Xe a =) Vi-vat-V2 + N noi 5 j& cju) >> (V-va) [5 - 3 ] +5V, -o [t-j 4 j -1)

=> (V-V2) ( 8- 5j) +jiov, zo >>|(875j]v+ C5j-8) Va=0 - a Apply Kul in mode (N2), following squette y KuL in wing equation nocle (N vi-V2vi-uz - V2-002 - v 518 XL2+R3 + 12 +112 L-30° =) (V-V2) (8-5j) - V2 Lio 5+8 j => W-V2) (8-5;) = 5V2+ (5+8;]V₂ -12 230 (5+81) uo 565+8j) -YU, -V2) (10-25j +64j+40) = 8 [Qvz (1048j) -12 ] =;-V2) (980+ 39j) - Va[ 20+84j) – 96 LocSaoji) -> (80434;) ¥I +(-160–103) -96130* (5+:1]

(80739j) Vit C-160-j103) U2 = 6-799.60 – j425:11) Svi Writing equation ③ & in matrix format (d) ( 8+5; 5;-8 +1425.1 80+30) -(16071033) AX-B ] -semear form, Solution to the equation is TX = A'B.] " l in Matlab & Let us solve x = fond .Celt) = ?

%MATLAB CODE

clear all
clc
Xc=4/i;
A=[8+5*i 5*i-8;80+39*i -(160+103*i)]; %Create A matrix
B=[0;-7.9969e+02 - 4.2511e+02i];
X= inv(A)*B; % Multiplying A inverse with B to get solution
V1=X(1);
V2=X(2);
Ic=V1/Xc;
display('Amplitude of Ic(t)')
Amp=abs(Ic)
display('Phase angle of Ic(t) in degrees')
phi=angle(Ic)*180/pi
syms amp_ic ph_ic f_ic
amp_ic = Amp
ph_ic = phi
f_ic =400

Command Window Amplitude of Ic(t) Amp = 1.2466 Phase angle of Ic(t) in degrees phi = -6.3520 amp_ic = 1.2466 ph_ic = -6.3520 f_ic = 400

ce) Ic= ju, 14-98651-96.352 Ic= 1.2466 / -6.352. Icct) = 1,2466 cos (400t – 6.352°)