SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self-paced Course MODULE 26 APPLICATIONS TO ELECTRICAL CIRCUITS Module Topics 1. Complex numbers and alternating currents 2. Complex impedance 3. Differential equations for RLC circuits 4. Forced oscillations and resonance 5. Complex solutions of differential equations 6. Phasors A: Work Scheme based on JAMES (FIFTH EDITION) 1. The first topic uses results from module 22 on complex numbers. Before starting on this topic you should make sure that you understand the relevant topics. If you need to revise this material look at the module again or study section 3.3 of James. 2. Turn to p.225 and study section 3.6 on alternating currents in electrical circuits. Work through Example 3.32. The key results of this section are as follows: When an alternating current with frequency ω/2π flows in a circuit it is given by Similarly we have i = Isinωt = Im ( Ie jωt). Icosωt = Re ( Ie jωt) The corresponding voltage depends on the resistance, capacitance and inductance of the circuit. However these devices not only determine the magnitude of the voltage but also change the phase of the voltage relative to the current. We may therefore calculate the voltage by multiplying Ie jωt by a complex number Z and then looking at the imaginary part. Note using a complex number deals with the change of phase since Im(e jφ e jωt ) = Im(e j(ωt+φ) ) = sin(ωt+φ), and Re(e jφ e jωt ) = Re(e j(ωt+φ) ) = cos(ωt+φ). The complex number Z is called the complex impedance of the device. The formula for Z for various devices is given as follows: { R for a resistor Z = j ωc for a capacitor jωl for an inductor For a simple RLC circuit (with these devices in series) R L C Fig. 1. 1
we simply take the sum of these terms so that Using this result the actual voltage is given by Z = R+jωL j ωc v = Im ( IZe jωt) = I Z sin(ωt+φ) where Z is the impedance of the circuit and the phase φ is given by the argument of Z. Note: it is often convenient to introduce the reactance S given by So that S = ωl 1 ωc Z = R+jS One example of the use of complex impedance is given in Example 3.32. Here is another example: Example A: Calculate the complex impedance of the element of the circuit shown below with a resistor of R = 20Ω and a capacitor of C = 212 µf, when an alternating current of frequency 50 Hz flows. Use this to find the impedance and the phase. Fig. 2. The complex impedance is the sum of the individual impedance s. Thus Z = R j ωc Here R = 20Ω, ω = 2π 50rad s 1 and C = 212µF, so that 1/Cω = 15 Z = 20 15j Hence the impedance is Z = (15) 2 +(20) 2 = 25Ω and the phase is φ = tan 1 ( 15/20) = 0.64 rad. ***Do Exercises 3.6.1 on p.227*** 3. Turn to p.794 and study section 10.2.4 on Simple Electrical Circuits. We again summarise the key results: The relationship between voltage V, current i and resistance R for a pure resistor is given by V = ir Similarly for a pure capacitor of capacitance C with charge q we have Finally for a pure inductor of inductance L we have V = q C V = L di dt 2
If we now consider the circuit shown below A B Fig. 3. then when the switch is in contact with B then by Kirchhoff s voltage law the total potential difference around the circuit must be zero so we have L di dt +Ri+ q C = 0 (1) The principle of conservation of charge tells us that the current flowing is equal to the rate of change of charge so that i = dq (2) dt Substituting for i (and its time derivative) in (1) using (2) gives L d2 q dt 2 +Rdq dt + 1 C q = 0 (3) This is a differential equation for the charge q. However it is more convenient to work with the current i and differentiating (3) with respect to time gives L d2 i dt 2 +Rdi dt + 1 C i = 0 (4) Alternatively we can use V = q/c in equation (3) to obtain an equation for the voltage LC d2 V dt 2 +RCdV dt +V = 0 (5) We have therefore shown that a simple analysis of a RLC circuit gives rise to a second order (homogeneous) linear differential equation with constant coefficients. Example B: A series circuit (as in Fig. 3) contains an inductor for which L = 1 H, a resistor for which R = 1 kω, and a capacitor for which C = 6.25 µf. The capacitor holds a charge of 1.5 10 3 C, at time t = 0 a switch is moved from A to B and the capacitor discharges through the circuit. Find q and i as functions of t. The differential equation to be solved is d 2 q dt 2 +103dq dt + 1 6.25 10 6q = 0 This is a differential equation with constant coefficients so we look at the auxiliary equation m 2 +1,000m+160,000 = 0 3
which has roots m 1 = 200 and m 2 = 800. Hence q = Ae 200t +Be 800t ( ) The initial charge is q 0 = 1.5 10 3 so that A+B = 1.5 10 3 (i) To find the other initial condition we differentiate (*) to obtain an equation for the current The initial current is zero so that we have i = 200Ae 200t 800Be 800t 200A 800B = 0 (ii) Solving (i) and (ii) for A and B we get A = 2 10 3, B = 5 10 4 Hence and q = 2 10 3 e 200t 5 10 4 e 800t i = 0.4e 200t +0.4e 800t ***Do Exercise A: A series circuit consists of an inductor for which L = 0.02 H and a capacitor for which C = 8 10 6 F. The capacitor holds a charge of 16 10 4 C and at time t = 0 a switch is closed allowing the capacitor to discharge through the circuit. Find the charge and the current in the circuit at time t. ***Do Exercise B: For the problem given above also find the voltage drop across the inductor and across the capacitor and show that the sum of these two voltages is zero in accordance with Kirchhoff s law. 4. Linear differential equations with constant coefficients are covered in module 13 (Differential Equations III) in order to be able to study oscillations in RLC circuits you need to make sure that you understand this material. If you need to revise this topic look again at module 13 or else turn to p.852 and read section 10.9 of J. Note that you will also need the material from section 10.9.3 which deals with the inhomogeneous equations with constant coefficients. ***Do Exercise 55 on p.857 and Exercise 63(c),(f) on p.864*** 5. Read section 10.10.4 on p.850 which deals with oscillations in electrical circuits. The key result is the following. Consider the RLC circuit shown below V i R L C V o Fig. 4. 4
Suppose that a voltage V i (t) is applied across the input terminals then the voltage V o (t) across the output terminals is given by LC d2 V o dt 2 +RCdV o dt +V o = V i (t) (6) which is second order inhomogeneous differential equation for constant coefficients. In many cases the input signal is of the form V i = V cosωt (where V is a constant) so that the equation reduces to LC d2 V o dt 2 +RCdV o dt +V o = V cosωt (7) To solve this we need to find a complementary function and a particular integral. The complementary function is just a solution of the homogeneous equation (5) and in section 10.10.2 of J it is shown that in this case (since R and C are positive) this is always given by a decaying function of time. For this reason the complementary function is called the transient solution. In order to find the particular integral we look for a solution of the form P cosωt+qsinωt where P and Q are constants which depend upon ω. This can be written in the alternative form AV C cos(ωt+δ) where A(ω) is a constant that depends upon ω and δ gives the phase relative to the input. This is a sinusoidal signal which oscillates with amplitude AV/C. Note that unlike the transient solution this solution does not decay so that this solution gives the long term behaviour of the solution. For this reason it is called the steady-state response. A direct calculation shows that A = 1 ω(r 2 +S 2 ) 1/2 = 1 ω Z where R is the resistance, S is the reactance and Z is the impedance, and the phase is given by δ = tan 1 ( R S ) Therefore if a sinusoidal voltage is applied to the input terminals the voltage produced is also sinusoidal but with an amplitude (and phase) which depends on the frequency of the input. Such a device is called a filter. Note that the corresponding current is obtained by differentiating the voltage and multiplying by C so that the modulus of the output current is the modulus of the input voltage divided by the impedance. We now give an example of how equation (6) may be used to calculate the voltage in an alternating RLC circuit. Example C: An RLC-circuit as shown in Fig. 4 consists of a resistor with R = 11 Ω, an inductor with L = 0.1 H, and a capacitor with C = 10 2 F. An input voltage V i (t) = 1000cos400t is applied. Find the resultant steady state voltage. Find the amplitude of this voltage and the phase lag compared to the the voltage that produced it. The differential equation for V o (t) is given by 10 3d2 V o dt 2 +0.11dV o dt +V o = 1000cos400t To find the steady state solution we look for a particular solution of the form V o = P cos400t+qsin400t 5
Differentiating this and substituting into the equation we get (44Q 159P)cos400t (44P +159Q)sin400t = 1000cos400t Hence which has solution 159P +44Q = 1000 44P +159Q = 0 P = 5.84, Q = 1.62 so that V o = 5.84cos400t+1.62sin400t. It is more useful to write this in terms of modulus and phase as V o = 6.06cos(400t+0.27) So that the amplitude is 6.06 and the voltage is 0.27 in advance of the voltage that produced it (since δ is negative). ***Do Exercise C: An RLC-circuit as shown in Fig. 4 consists of a resistor with R = 100 Ω, an inductor with L = 0.1 H, and a capacitor with C = 10 5 F. An input voltage V i (t) = 100cos500t is applied. Find the resultant steady state voltage and current and the amplitude of this current. 6. Complex Solutions The following sections deal with complex solutions and phasors which are not described in J. As we have seen the equation for the output current is where V i (t) is the input voltage. Two cases of interest are L d2 q dt 2 +Rdq dt + 1 C q = V i(t) (8) and V i (t) = V cos(ωt) V i (t) = V sin(ωt) (a) (b) If we solve equation (8) with a complex source given by V i (t) = V(cos(ωt)+jsin(ωt)) = Ve jωt (c) then we obtain a complex solution q(t). However the real part of this is gives the solution corresponding to source (a), while the imaginary part of this solution gives the solution corresponding to source (b). It turns out that it is easier to solve (8) for the complex source (c) and then take the real part rather than solve for the real source (a). For the case of the complex source we want to solve L d2 q dt 2 +Rdq dt + 1 C q = Vejωt (9) The steady-state solution is given by the particular solution and we find this by looking for a solution of the form q(t) = KVe jωt where K is a complex constant to be determined. Differentiating this gives dq dt = jωkvejωt, and d 2 q dt 2 = ω2 KVe jωt 6
Substituting into (9) gives ( ω 2 L+jRω + 1 C )KVejωt = Ve jωt and hence using S = ωl 1, we must have ωc So that jω(r+js)k = 1 K = 1 jωz where Z is the complex impedance. Thus the corresponding complex current is given by i c (t) = dq dt = V Z ejωt To find the actual (real) current we write the complex impedance in polar form as Z = Z e jθ where ) Z = R 2 +S 2, and θ = tan 1 ( S R Then the complex current produced is So that the real current produced is i c (t) = V Z ej(ωt θ) i(t) = Re(i c (t)) = V Z cos(ωt θ) To summarise an input voltage of Re(Ve jωt ) produces an output current of Re( V Z ejωt ). 7. Phasors If we are given an alternating current i(t) = I o cos(ωt + φ) in a circuit which is oscillating with angular frequency ω, then all we need to know to find the current is the quantity I o e jφ and the angular frequency ω. This information is provided by the current phasor since the actual current is given by Ĩ(jω) = I o e jφ i(t) = Re(Ĩejωt ) = Re(I o e j(ωt+φ) ) In the same way the voltage v(t) = V 0 cos(ωt+θ) is determined by the voltage phasor Ṽ(jω) = V 0 e jθ In the example of the RLC-circuit considered in section 6 we see that Ĩ(jω) = Ṽ(jω) Z(jω) 7
or Ṽ(jω) = Z(jω)Ĩ(jω) where Ṽ(jω) is the phasor for the input voltage, Ĩ(jω) is the phasor for the output current and Z(jω) is the complex impedance of the circuit. Note by taking the modulus of this expression we immediately see that the amplitude of the input voltage is the impedance times the amplitude of the output current. By looking at the argument we also see that the phase of the output current is argz behind that of the input voltage. Example D: Use the phasor method to calculate the amplitude and phase of the output current for the circuit considered in Example C. Use this to compute the modulus of the output voltage. In this example R = 11, ω = 400 and C = 10 2. Hence S = ωl 1 ωc = 159/4 and hence the complex impedance is Z = 1 4 (44+159j). Thus Z = R 2 +S 2 = 41.243, and argz = tan 1 (159/44) = 1.3rad Hence the amplitude of the current is 1000/41.243 = 24.25 and the phase is 1.3 rad. Hence the output current is i(t) = 24.25cos(400t 1.3) To find the voltage we integrate and divide by C. Hence the modulus of the output voltage is given by 24.25 400 0.01 = 6.06 in agreement with the previous answer. Note the phase difference of the voltage is π/2 1.3 = 0.27 again in agreement with the previous calculation (the extra π/2 coming from the fact that a cos becomes a sin on integration). B: Work Scheme based on STROUD (SIXTH EDITION) This module is not covered by S., so work through A: Work scheme based on J, presented above. 8
Specimen Test 26 1. An alternating current i = 2 sin 100t flows through a circuit consisting of a resistor with resistance R = 50 Ω, and an inductor with inductance L = 0.5 H, as shown below R L (i) Calculate the complex impedance Z of the circuit. (ii) Calculate the amplitude V of the voltage. (iii) Show that the voltage may be written in the form v(t) = V sin(100t+φ) where V is the amplitude and φ is the phase which you should calculate. 2. The complex impedance of two circuits in parallel with complex impedances Z 1 and Z 2 is given by 1 Z = 1 Z 1 + 1 Z 2 If Z 1 = 1+2j and Z 2 = 2 j calculate Z in the form Z = a+bj and use this to calculate the (real) impedance of the circuit. 3. A simple RLC-circuit shown below contains and inductor with L = 1 H, a resistor for which R = 10 3 Ω, and a capacitor for which C = 4 10 6 F. With the switch in position A, the battery maintains a charge of 5 10 4 C in the capacitor. At time t = 0 the switch is moved to B and the capacitor discharges through the circuit. A B (i) Find the differential equation for the charge q(t) and find the general solution. (ii) Find the expression for the corresponding general solution for the current i(t). (iii) Hence find the solutions q(t) and i(t) which satisfy the initial conditions for this circuit at t = 0. 9
4. The RLC-circuit shown below V i R L C V o has R = 8 Ω, L = 0.5 H, C = 0.1 F and input voltage given by V i = 100sin2t. (i) Find the differential equation satisfied by the output current i o (t). (ii) Find a particular solution for the differential equation and hence find the steady state output current and calculate its amplitude. (iii) Confirm your answer for the amplitude of the output current by using the method of phasors, and calculate the phase of the current relative to the input voltage. 10