A sinusoidal AC voltage source is applied to a circuit. There are two impedances
ID: 2265885 • Letter: A
Question
A sinusoidal AC voltage source is applied to a circuit. There are two impedances, Zy and Z2, in the circuit, one is a resistor and an inductor connected in parallel, with R, = 8 , L-5x103 H. The other is a resistor and a capacitor connected in series, with R2 = 6 , C= 10 F. (1) Calculate the total impedance, total current (instantaneous and RMS), voltage drop (instantaneous and RMS) and current (instantancous and RMS) on each impedance (Z1 and Z2), instantaneous power, apparent power, active power, reactive power, and power factor of each impedance (Z1 and Z2): The two impedances are connected in parallel, and The two impedances are connected in series. a. b. SOLVE IN MATLAB USING SCRIPT DONT TRY TO SOLVE IF YOURE NOT USING MATLAB. NEED TO SEE HOW TO WRITE CODEExplanation / Answer
//for series use this code
Vm=(120*sqrt(2));
Vrms=(Vm/sqrt(2));
syms t;
V=(Vm*cos((120*pi*t)+30));
R1=8;
L=(5*(10.0e-03));
f=60;
XL=(2*pi*f*L);
R2=6;
C=(10*(1.0e-06));
XC=(1/(2*pi*f*C));
Z1=(R1*(XL)*i)/(R1+((XL)*i));
Z2=(R2+((XC)*i));
Z=Z1+Z2;
Iinst=(V/Z);
Irms=((Vrms/Z);
Vdropinst=(Iinst*Z);
Vdroprms=(Irms*Z);
I1inst=Iinst;
I2inst=Iinst;
I1rms=Irms;
I2rms=Irms;
Pinst=(V*Iinst);
Paprnt=(Vrms)*Irms;
pf=cos(angle(Z));
k=angle(Z);
Pactv=(Vrms*Irms*cos(k));
Prectv=(Vrms*Irms*sin(k));
Pf1=cos(angle(Z1));
Pf2=cos(angle(Z2));
//for parallel use this code
Vm=(120*sqrt(2));
Vrms=(Vm/sqrt(2));
syms t;
V=(Vm*cos((120*pi*t)+30));
R1=8;
L=(5*(10.0e-03));
f=60;
XL=(2*pi*f*L);
R2=6;
C=(10*(1.0e-06));
XC=(1/(2*pi*f*C));
Z1=(R1*(XL)*i)/(R1+((XL)*i));
Z2=(R2+((XC)*i));
Z=(Z1*Z2)/(Z1+Z2);
Iinst=(V/Z);
Irms=((Vrms/Z);
Vdropinst=(Iinst*Z);
Vdroprms=(Irms*Z);
I1inst=(Vinst/Z1);
I2inst=(Vinst/Z2);
I1rms=Irms;
I2rms=Irms;
Pinst=(V*Iinst);
Paprnt=(Vrms)*Irms;
pf=cos(angle(Z));
k=angle(Z);
Pactv=(Vrms*Irms*cos(k));
Prectv=(Vrms*Irms*sin(k));
Pf1=cos(angle(Z1));
Pf2=cos(angle(Z2));