Numerical Analysis of Circuits Containing Resistors, Capacitors and Inductors
NUMERICAL ANALYSIS OF CIRCUITS CONTAINING RESISTORS, CAPACITORS AND INDUCTORS
Numerical methods are often used to study mechanical systems. However, these same methods can be employed to investigate the response of circuits containing resistors, capacitors and inductors. A wide range of phenomena can be studied such as the frequency response of filters and tuned circuits; resonance, damped and forced oscillations; voltage, current, power, energy and phase relationships.
Resistors, capacitors and inductors are basic components of circuits. These components are connected to a source of electrical energy. A simple model for the source of electrical energy is to consider it to be a voltage source called the emf vS and a series resistance called the internal resistance r. The potential difference applied to a circuit is called the terminal voltage v. If a current is supplied from the electrical energy source to a circuit, the terminal voltage is
v = vS – i r
In the computer modelling of circuit behaviour, the effects of internal resistance of the source can be considered. Suitable electrical sources include step, on/off/on, sinusoidal and pulsed functions.
For a resistorR, the voltage vR across it and the current iR through it are always in phase and related by the equation
vR(t) = R i(t)
A capacitor consists of two metal plates separated by an insulating material. When a potential difference vC exists across a capacitor, one plate has a positive charge +q and the other plate a charge –q. The charge q is proportional to the potential difference between the plates of the capacitor, the constant of proportionality is called the capacitance (farad F).
q(t) = CvC(t)
Differentiating both sides of this equation with respect to t and using the fact that
i = dq/dt we get dvC/dt = iC / C. We can write this equation in terms of differences rather than differentials
Thus, a capacitor will resist changes in the potential difference across it because it requires a time t for the potential difference to change by vC. The potential difference across the capacitor decreases when it discharges and increases when charging. The larger the value of C, the slower the change in potential. If t is “small”, then to a good approximation, the potential difference across the capacitor at time t is
An inductor can be considered to be a coil of wire. When a varying current passes through a coil, a varying magnetic flux is produced and this in turn induces a potential difference across the inductor. This induced potential difference opposes the change in current. The potential difference across the inductor is proportional to the rate of change of the current, where the constant of proportionality L is known as the inductance of the coil (henries H) and is given by the equation
vL = L di/dt
This equation can be expressed in terms of differences,
iL(t) = vL(t) t / L
Thus, an inductor will resist changes in the current through it because it requires a time t for the current to change by i. If t is “small”, then to a good approximation, the current through the inductor at time t is
SERIES RC CIRCUIT
Kirchhoff’s Laws with the equation
can be used to compute the response of a series RC circuit for different applied emfs. The first step is calculate the applied emf vs and then specify the initial values for the variables for voltage v, current i, powers p and energies u.
vS(0)
vC(0) = 0 assume capacitor is initially discharged
i(0) = { vS(0) – vC(0) } / R
vR(0) = i(0) R
pS(0) = vS(0) i(0) pR(0) = vR(0) i(0) pC(0) = vC(0) i(0)
uS(0) = 0 uR(0) = 0 uC(0) = 0
Values at all later times are found by implementing the routine in the order shown
For accurate results the time increment t should be chosen so that
tR C where R C is the capacitive time constant.
MATHLAB FILE FOR SERIES RC CIRCUIT
%m260ah.m
%1 feb 01
%numerical analysis of an RC circuit
tic
R = 500;
C = 1e-6;
tau = R*C; %time constnat
dt = 1e-5; %time increment dt < RC;
%construction of a rectangular pulse
num = 500; %number of points 1 to 500
T = 3; %number of periods
ontime = 50; %percentage of time pulse is on - must be less than 100;
numT = round(num/T-0.5); %number of points for period
ton = round(numT*ontime/100); %on time expressed as number of points
toff = numT - ton; %off time expressed as number of points
vs = zeros(num,1); %set voltages to zero
vmax = 2; %set max on voltage
for c = 1 : T
vs(toff+numT*(c-1):toff+numT*(c-1)+ton) = vmax;
end
%zero variables
t = zeros(num,1);
i = zeros(num,1);
vr = zeros(num,1);
vc = zeros(num,1);
ps = zeros(num,1);
pr = zeros(num,1);
pc = zeros(num,1);
us = zeros(num,1);
ur = zeros(num,1);
uc = zeros(num,1);
%calculations
time = 0:dt:dt*(num-1);
for c = 2 : num
vc(c) = vc(c-1)+i(c-1)*dt/C;
i(c)= (vs(c)-vc(c))/R;
vr(c) = i(c)*R;
end
ps = vs .* i;
pc = vc .* i;
pr = vr .* i;
for c = 2 : num
us(c) = us(c-1)+ps(c)*dt;
uc(c) = uc(c-1)+pc(c)*dt;
ur(c) = ur(c-1)+pr(c)*dt;
end
%graphing
figure(1);
plot(time',vs);
axis([0 dt*num -2.5, 2.5]);
title('RC CIRCUIT');
xlabel('time t (s)');
ylabel('voltage (V)');
grid on;
hold on;
plot(time',vc,'m');
plot(time',vr,'k');
legend('Vs','Vc','Vr');
figure(2);
plot(time',ps);
title('RC CIRCUIT');
xlabel('time t (s)');
ylabel('power P (W)');
grid on;
hold on;
plot(time',pc,'m');
plot(time',pr,'k');
legend('Ps','Pc','Pr');
figure(3);
plot(time',us);
title('RC CIRCUIT');
xlabel('time t (s)');
ylabel('energy U (J)');
grid on;
hold on;
plot(time',uc,'m');
plot(time',ur,'k');
legend('Us','Uc','Ur');
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PARALLEL TUNED LC CIRCUIT
Kirchhoff’s Laws with the equations
can be used to compute the response of a parallel tuned LC circuit for difference applied emfs. The first step is calculate the applied emf vS and then specify the initial values for the variables for voltages v, current i, powers p and energies u.
vS(0) = vR(0) vp(0) = 0 assume capacitor is initially discharged
iR(0) = vR(0) / R iRL(0) = vp(0) / RL = 0 iL(0) = 0 iC(0) = iR(0)
pS(0) = vS(0) i(0) pR(0) = vR(0) i(0) pC(0) = vC(0) i(0) pRL(0) = vRL(0) i(0)
uS(0) = 0 uR(0) = 0 uC(0) = 0 uR(0) = 0
Values at all later times are found by implementing the routine in the order shown
For accurate results the tine increment t should be chosen so that
t < 1 / f where f is the frequency of the emf.
MATLAB FILE FOR PARALLEL TUNERD CIRCUIT
%m260aj.m
%2 feb 00
%parallel tuned LC circuit with load resistor
%numerical analysis - leap frog method
%sinusoidal emf
tic
clear;
%data
R = 1000;
RL = 1e3;
C = 1e-8;
L = 3.8e-3;
vSo = 10; %emf amplitude
f = 25e3; %emf frequency
period = 1/f;
tmax = 4/f; %max time interval for calculations and graphs
num = 500; %number of data points 1 to 500
dt = tmax / (num-1);
t = 0 : dt : tmax;
vS = vSo .*sin((2*pi*f).*t);
%initialise values
vR(1) = vS(1);
vp(1) = 0; %voltage across parallel combination
iR(1) = vR(1)/R;
iC(1) = iR(1);
iL(1) = 0;
iRL(1) = 0;
%calculations
for c = 2 : num
vp(c) = vp(c-1) + iC(c-1)*dt/C;
vR(c) = vS(c) - vp(c);
iR(c) = vR(c)/R;
iRL(c) = vp(c)/RL;
iL(c) = iL(c-1) + (vp(c-1))*dt/L;
iC(c) = iR(c)-iRL(c)-iL(c);
end
pRL = vp .*iRL;
pR = vR .*iR;
pS = vS .*iR;
pC = vp .*iC;
pL = vp .*iL;
figure(1);
plot(t,vp);
title('Parallel LC Tuned Circuit');
xlabel('time t (s)');
ylabel('potential difference V (V)')
hold on;
plot(t,vS,'m');
figure(2);
plot(t,pRL,'k');
title('Parallel LC Tuned Circuit');
xlabel('time t (s)');
ylabel('power P (W)')
hold on;
plot(t,pS,'b');
plot(t,pR,'m');
plot(t,pC,'g');
plot(t,pL,'c');
legend('RL','S','R','C','L');
%accuracy of model dt < period
fo = 1/(2*pi*sqrt(L*C))
period
dt
%max power transferred to load Thevenin circuit theorem
w = 2*pi*f;
z1 = R;
z2 = -j/(w*C);
z3 = j*w*L;
z4 = 1/(1/z2+1/z3);
zth = 1/(1/z1+1/z4);
zthreal = real(zth)
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260ah.doc 16/12/18 8:27 PM