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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 resistor R, 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) = C vC(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 260ah.doc 15/05/17 3:51 AM 1 vC (t ) iC (t ) t / C 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 vC (t ) vC (t t ) iC (t t ) t / C 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 iL (t ) iL (t t ) vL (t t ) t / L 260ah.doc 15/05/17 3:51 AM 2 SERIES RC CIRCUIT R vs C Kirchhoff’s Laws with the equation vC (t ) vC (t t ) iC (t t ) t / C 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 vC (t ) vC (t t ) iC (t t ) t / C i (t ) vS (t ) vC (t ) / R vR (t ) i (t ) R pS (t ) vS (t ) i (t ) pC (t ) vC (t ) i (t ) pR (t ) vR (t ) i (t ) i t 1 uS (t ) pSi t pS (t ) t i 1 i t 1 uC (t ) pCi t pC (t ) t i 1 i t 1 uR (t ) pRi t pR (t ) t i 1 For accurate results the time increment t should be chosen so that t << R C where R C is the capacitive time constant. 260ah.doc 15/05/17 3:51 AM 3 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; dt = 1e-5; %time constnat %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); vmax = 2; %set voltages to zero %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; 260ah.doc 15/05/17 3:51 AM 4 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'); toc 260ah.doc 15/05/17 3:51 AM 5 RC CIRCUIT 2.5 2 1.5 1 voltage (V) 0.5 0 -0.5 -1 -1.5 Vs Vc Vr -2 -2.5 0 0.5 1 1.5 2 -3 8 2.5 time t (s) 3 3.5 4 4.5 5 -3 x 10 RC CIRCUIT x 10 Ps Pc Pr 6 power P (W) 4 2 0 -2 -4 -6 0 0.5 260ah.doc 15/05/17 3:51 AM 1 1.5 2 2.5 time t (s) 3 3.5 4 4.5 5 -3 x 10 6 -6 9 RC CIRCUIT x 10 Us Uc Ur 8 7 6 energy U (J) 5 4 3 2 1 0 -1 0 0.5 260ah.doc 15/05/17 3:51 AM 1 1.5 2 2.5 time t (s) 3 3.5 4 4.5 5 -3 x 10 7 PARALLEL TUNED LC CIRCUIT R vS C L RL Kirchhoff’s Laws with the equations vC (t ) vC (t t ) iC (t t ) t / C iL (t ) iL (t t ) vL (t t ) t / L 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 iR(0) = vR(0) / R assume capacitor is initially discharged 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) uS(0) = 0 uR(0) = 0 uC(0) = 0 pC(0) = vC(0) i(0) pRL(0) = vRL(0) i(0) uR(0) = 0 Values at all later times are found by implementing the routine in the order shown vp (t ) vp (t t ) iC (t t ) / C vR (t ) vS (t ) vp (t ) iR (t ) vR (t ) / R iRL (t ) vp (t ) / RL iL (t ) iL (t t ) vp (t t ) t / L iC (t ) iR (t ) iRL (t ) iL (t ) pS (t ) vS (t ) iR (t ) pR (t ) vR (t ) i (t ) pRL (t ) vRL (t ) i (t ) pC (t ) vC (t ) i (t ) pL (t ) vL (t ) i (t ) For accurate results the tine increment t should be chosen so that t << 1 / f where f is the frequency of the emf. 260ah.doc 15/05/17 3:51 AM 8 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; f = 25e3; period = 1/f; %emf amplitude %emf frequency 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; 260ah.doc 15/05/17 3:51 AM 9 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 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) Thevenin circuit theorem toc Parallel LC Tuned Circuit 10 8 6 potential difference V (V) 4 2 0 -2 -4 -6 -8 -10 0 0.2 260ah.doc 15/05/17 3:51 AM 0.4 0.6 0.8 time t (s) 1 1.2 1.4 1.6 -4 x 10 10 Parallel LC Tuned Circuit 0.08 RL S R C L 0.06 power P (W) 0.04 0.02 0 -0.02 -0.04 0 0.2 260ah.doc 15/05/17 3:51 AM 0.4 0.6 0.8 time t (s) 1 1.2 1.4 1.6 -4 x 10 11