loop currents

Question: 4.1
(¦(¦(18&-4)&-8@¦(¦(-4@-8)&¦(12@-2))&¦(-2@25))) (¦(I1@I2@I3))=(¦(10@0@0))

% this program determines the current
% flowing in a resistor RB and power supplied by source
% it computes the loop currents given the impedance
% matrix Z and voltage vector V
% Z is the impedance matrix
% V is the voltage matrix
% initialize the matrix Z and vector V
Z = [18 -4 -8;
-4 12 -2;
-8 -2 25];
V = [10;
0;
0];
% solve for the loop currents
I = inv(Z)*V;
% current through RB is calculated
IRB = I(3) – I(2);
fprintf(‘the current through R is %8.3f Amps n’,IRB)
diary
the current through R is -0.030 Amps
Question 4.2

 
(¦(¦(3&0@0&11)&¦(-2&0@-6&-5)@¦(-15&-6@0&-1)&¦(31&-10@-2&3))) (¦(¦(V1@V2)@¦(V3@V4)))=(¦(¦(12@-90)@¦(120@36)))

% program computes the nodal voltages
% given the admittance matrix Y and current vector I
% Y is the admittance matrix and I is the current vector
% initialize matrix y and vector I using YV=I form
Y = [ 3 0 -2 0;
0 11 -6 -5;
-15 -6 31 -10;
0 -1 -2 3];
I = [12;
-90;
120;
36];
% solve for the voltage
fprintf(‘Nodal voltages V1, V2, V3 and V4 are n’)
v = inv(Y)*I
diary
Nodal voltages V1, V2, V3 and V4 are

v =

40.0000
50.7857
54.0000
64.9286

 
Question 4.3

 
(¦(¦(2&10@2&2)&¦(-12&0@10&6)@¦(0&0@0&12)&¦(-4&4@-14&6))) (¦(¦(I1@I2)@¦(I3@I4)))=(¦(¦(10@-48)@¦(-30@-16)))
% This program determines the power dissipated by
% 8 ohm resistor and current supplied by the
% 10V source
%
% the program computes the loop currents, given
% the impedance matrix Z and voltage vector V
%
% Z is the impedance matrix
% V is the voltage vector
% initialize the matrix Z and vector V of equation
% ZI=V
Z = [2 10 -12 0;
2 2 10 6;
0 0 -4 4;
0 12 -14 6];
V = [10 -48 -30 -16]’;
% solve for loop currents
I = inv(Z)*V;
% the power dissipation in 4 ohm resistor is P
P = 4*I(4)^2;
% print out the results
fprintf(‘Power dissipated in 4 ohm resistor is %8.2f Wattsn’,P)
% current is calculated
I = I(4);
fprintf(‘the current is %8.3f Amps n’,I)
diary
Power dissipated in 4 ohm resistor is 208.55 Watts
the current is -7.221 Amps

4.4
ohm resistor

(¦(¦(10&2)&2@¦(¦(18@8)&¦(8@12))&¦(8@3))) (¦(I1@I2@I3))=(¦(15@15@0))
% This program determines the power dissipated by
% 8 ohm resistor and current supplied by the
% 10V source
%
% the program computes the loop currents, given
% the impedance matrix Z and voltage vector V
%
% Z is the impedance matrix
% V is the voltage vector
% initialize the matrix Z and vector V of equation
% ZI=V
Z = [10 2 2 ;
18 8 8 ;
8 12 3 ];
V = [15 15 0 ]’;
% solve for loop currents
I = inv(Z)*V;
% the power dissipation in 15 voltag is P
P = 15*I(1);
% print out the results
fprintf(‘Power dissipated in 4 ohm resistor is %8.2f Wattsn’,P)
% current is calculated
I = I(1);
fprintf(‘the current is %8.3f Amps n’,I)
diary
Power dissipated in 4 ohm resistor is 30.68 Watts
the current is 2.045 Amps

10kO and 40kO are parallel
R=(10k×40k)/(10k+40k)=8k

VC(0)=0V
VC(8)=Vab=8k/(20k+8k)×30=60/7V
VC(t)=VC(8)+( VC(0)- VC(8))e^((-t)/T)
Req=(8k×20k)/(8k+20k)=5.714kO
T=60/7=5.714k ×1×?10?^(-3)=5.714s
Between 0=t=5
t=0
VC(t)= 60/7+( 0- 60/7)e^((-5)/5.714)
VC(t)=4.998˜5V
Using KVL, we get
L (di(t) )/dt+Ri(t)=VS
i(t)=Vs/R(1-e^(Rt/L))
Between 0 to 10 ms
i(t)=9/16(1-e^((16×10×?10?^(-3))/4))
i(t)=9/16(1-e^0.04)
i(t)=-0.02295606048
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