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Circuit Theory March 28, 2018

V-I Relationship for Resistors If there is a current of the form i = Im cos(ωt + θi ), then the voltage for the resistor will be v = Ri = R[Im cos(ωt + θi )] = RIm cos(ωt + θi ) If we applied a phasor transform to the above equation, we would have that V = RIm ej θi = RIm ∠(θi ) = RI → V = RI Note that in the above equation, we have replaced the current amplitude Im and the angle ∠(θi ) with the phasor of the current, I . In phasor domain, the current and the voltage are ”in-phase”.

V-I Relationship for Inductors If there is a current of the form i = Im cos(ωt + θi ), using the fact that the voltage for di an inductor v = L , we can write dt v=L

di d(Im cos(ωt + θi )) =L = −ωLIm sin(ωt + θi ) = ωLIm cos(ωt + θi − 90◦ ) dt dt

In the above equation, we have used the fact that sin(α) = cos(α − 90◦ ). If we do a phasor domain transform on the above equation, we have that V = −ωLIm ej(θi −90

◦)

= −ωLIm ej θi e−j90 = jωLIm ej θi = jωLI ◦

We say that voltage is ”leading” the current by 90 degrees. Similarly, we can also say that current is ”lagging” voltage by 90 degrees. 90◦ corresponds to one fourth of a period ( T4 ).

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V-I Relationship for Capacitors If we have a voltage of the form v = Vm cos(ωt + θv ), then by using the relation for dv current and voltage in time domain i = C , we have that dt i=C

d(Vm cos(ωt + θv )) dv =C = −CωVm sin(ωt + θv ) = −CωVm cos(ωt + θv − 90◦ ) dt dt

By going into the phasor domain, we will have that I = −CωVm ej θv e−j90 = jωCVm ej θv = jωCV ◦

We can rewrite the above equation and get V =

I I Im ∠(−90◦ ) = −j = ωC jωC ωC

We say that the voltage lags the current by 90◦ . We can also say that the current leads the voltage by 90◦ .

General V-I Relationship for R,L,C For a passive circuit element, we can write the following relationship in the phasor domain. V =Z ·I Where only V and I are phasors.Z represents the impedance of the circuit element and it is measured in Ohms (Ω).Note that Z can refer to any series or parallel combination of R, L and\or C. Although Z is a complex number, it is NOT a phasor. The imaginary part of the impedance (ℑ{Z }) is called reactance. The reactance of a resistor is 0. The reactance of an inductor is ωL. −1 The reactance for a capacitor is ωC . Example: An inductor is connected to a current source that produces an alternating current given by the relation i = 10 cos(10000t + 30◦ )mA. If L = 20mH find: a) The reactance b) The impedance c) Voltage of the inductor d) Steady state voltage of the inductor

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Answers: a) Recall that the reactance is only the imaginary part of the impedance. Since the impedance for an inductor is purely imaginary, we then have that ℑ{ZL } = ωL = 10000 · 20 · 10−3 = 200Ω b) The impedance will be jωL = j200Ω (note the difference between the value of the impedance and the value of the reactance) c) We must first find the phasor of the current. p(i) = p(10 cos(10000t + 30◦ ) = I = 10∠(30◦ )mA We then use the V-I relationship in phasor domain to find the voltage of the inductor. V = ZI = jωLI = j200 · 10ej30 · 10−3 = 2ej120 V ◦

◦

d) To revert back to time domain, we do an inverse phasor transform on the voltage phasor. ◦ p−1 (V ) = ℜ(2ej120 ej10000t ) = 2 cos(10000t + 120◦ )V

Kirchoff’s Laws in Frequency Domain Kirchoff’s Voltage Law for Phasors The sum of all voltage phasors in a closed loop is 0. V1 + V2 + V3 + ... + Vn =

n X

Vk = 0

k=0

The proof for this is the following. First, write the sum of all the steady state voltages in a loop (KVL). v1 +...+ vn = Vm1 cos(ωt+θ1 )+Vm2 cos(ωt+ θ2 )+Vm3 cos(ωt+θ3 )+ ...Vmn cos(ωt+θn ) = n X

Vmk cos(ωt + θk ) = 0

k=0

This is complex to analyse in time domain, so we do a phasor transform. ℜ{Vm1 cos(ωt + θ1 )} + ℜ{Vm2 cos(ωt + θ2 )} + ... + ℜ{Vmn cos(ωt + θn )} = ℜ{Vm1 ejωt ej θ1 } + ℜ{Vm2 ej ωt ej θ2 } + ... + ℜ{Vmn ej ωt ej θn } =

n X k=0

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ℜ{Vk ejωt ej θk } = 0

Note that the function ℜ{} that takes a complex number is linear, so we can unite all terms under the same ℜ{}.(equivalently, we’re pulling out the ℜ function from underneath the sum symbol) n X ℜ{Vm1 ejωt ej θ1 + Vm2 ej ωt ej θ2 + Vm3 ej ωt ej θ3 + ... + Vmn ej ωt ej θn } = ℜ{ Vk ejωt ej θk } = 0 k=0

Now we factor out ejωt . ℜ{(Vm1 ejθ1 + Vm2 ej θ2 + Vm3 ej θ3 + ... + Vmn ej θn )ej ωt } = ℜ{ej ωt

n X

Vk ejθk } = 0

k=0

ejωt can never be 0. In fact, its modulus is always 1 (can you see why? try to prove this for yourselves!). It remains that the only terms that can be 0 are the phasors. V1 + V2 + V3 + ... + Vn =

n X

Vk = 0

k=0

Kirchoff’s Current Law for Phasors The sum of all current phasors in a node is 0. I1 + I2 + I3 + ...In =

n X

Ik = 0

k=0

The proof is very similar to the proof for the voltage phasors. Write the sum of all currents for a node in time domain, do a phasor domain transformation then use the fact that ℜ is linear. Try to prove it for yourselves!

Series Impedances

We denote the voltage between a and b as Vab . Note that the current is the same for all impedances. Vab = IZ1 + IZ2 + ...IZn = I(Z1 + Z2 + ...Zn ) = I(

n X k=0

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Zk )

We can divide the 2 parts of the equation by I and get X Vab = Z1 + Z2 + ...Zn = Zk I n

k=0

The phasor of the voltage divided by the phasor of the current is the equivalent impedance Vab = Zeq . I n X Zk Zeq = k=0

This is to say that we can replace the impedances in series with an equivalent impedance that is the sum of all impedances. Example:

A 90Ω resistor in series with a 32mH inductor and a 5µF capacitor are connected to a voltage source v(t) = 750 cos(5000t + 30◦ ). Find a) The frequency domain equivalent circuit b) The steady state current Answers: Every time you build the phasor domain equivalent circuit, you must follow 3 steps. 1) Calculate all voltage and current phasors. 2) Calculate all impedances (you must know ω to do this) 3) Build the equivalent phasor domain circuit. We have the steady state voltage of the source, so we will first write the phasor of the voltage. p(v) = V = 750∠30◦ Next, we have that the frequency is 5000 rad . sec ZR = R = 90Ω ZL = jωL = j5000 · 32 · 10−3 = j160Ω 1 106 = −j40Ω ZC = = −j 5000 · 5 jωC Since we have everything, we can now build the phasor domain equivalent circuit.

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b) All elements are in series, so there must be the same amount of current flowing through all of them. We have the voltage phasor and the impedances and from these 2 we can find the current phasor. First, find the equivalent impedance of the series elements. Zeq = ZR + ZL + ZC = 90 + 160j − 40j = 90 + 120j = 150ej53.13 Ω ◦

The current will be the voltage divided by the equivalent impedance. V 750ej30 ◦ I= = ◦ = 5∠(−23. 13 )A Zeq 150ej53.13 ◦

To find the steady state current, revert from phasor domain. i = p−1 (I) = 5 cos(5000t − 23.13◦ )A

Parallel Impedances

We begin by writing KCL. I = I1 + I2 + ... + In But each current is the voltage (which is equal for each branch) divided by the impedance of that branch. Vab Vab Vab Vab = + + ... Zeq Z1 Z2 Zn Divide by Vab

X 1 1 1 1 1 = = + + ... Zk Zeq Z1 Z2 Zn n

k=0

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The inverse of the impedance is called admittance and it is given by Y =

1 = G + jβ Siemens Z

G is the conductance β is susceptance. Having given these definitions , we can rewrite the above equation for parallel impedances in terms of admittances. Yab = Y1 + Y2 + Y3 + ...Yn =

n X

Yk

k=0

For a resistor, the admittance is G. 1 . and the susceptance is −1 For an inductor, the admittance is jωL ωL For a capacitor, the admittance is jωC and the susceptance is ωC . Example:

For this circuit, it is known that is = 8 cos(200000t)A.Find a) The phasor domain equivalent circuit b) The steady state voltage v and the 3 currents i1 , i2 and i3 . Answers: The phasor of the current is given by Is = 8∠(0◦ )A We find next the admittance of each branch. Y1 Y2 Y3

1 = 0.1S 10 6 − 8j 1 = = 0.06 − j0.08S = 6 + 8j 100 = jωC = j0.2S =

Yeq = Y1 + Y2 + Y3 = 0.16 + j0.12 = 0.2∠(36.87◦ )S At this stage, we can also take the inverse of the admittances and find the impedances to build the equivalent circuit in phasor domain.

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The equivalent impedance is the inverse of the equivalent admittance. 1 = 5∠(−36.87◦ )Ω Zeq = Yeq The voltage is the impedance multiplied by the current. V = Zeq I = 40∠(−36.87◦ )V Now we can find the current through each branch by dividing the voltage by the impedance (or multiplying by the admittance) 40∠(−36.87◦ ) = 4∠(36.87◦ )A 10 40∠(−36.87◦ ) = 4∠(−90◦ ) = −j4A = V Y2 = 6 + 8j 40∠(−36.87) = 8∠(53.13◦ )A = V Y3 = 5∠(−90◦ )

I1 = V Y1 = I2 I3

To find the steady state expressions for the currents and the voltage, we do inverse phasor transforms for all the current and voltage phasors. v = 40 cos(200000t − 36.87◦ )V i1 = 4 cos(200000t − 36.87◦ )A i2 = 4 cos(200000t − 90◦ )A i3 = 8 cos(200000t + 53.13◦ )A

Delta-to-Y:

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We can derive the Y equivalent for impedances in a similar manner to how we derived the Y equivalent for resistors. Zb Zc Za + Zb + Zc Zc Za Z2 = Za + Zb + Zc Za Zb Z3 = Za + Zb + Zc Z1 =

We can also convert any Y circuit into an equivalent ∆ circuit. Z1 Z2 + Z2 Z3 + Z3 Z1 Z1 Z1 Z2 + Z2 Z3 + Z3 Z1 Zb = Z2 Z1 Z2 + Z2 Z3 + Z3 Z1 Zc = Z3

Za =

Source Transformation

Example:

Using source transformations, find V0 . Answers: First we transform the voltage source in series with the inductor and resistor

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into a current source in parallel with the same impedance. The current source will be equal to the voltage of the source divided by the impedance 40 = 4 − 12jA I = 1+3j = 40(1−3j) 10

Next, we have 2 impedances in parallel, so we can find their equivalent impedance. (1 + 3j)(9 − 3j) Z1 Z2 = 1.8 + j2.4Ω = 10 Z1 + Z2

Zeq =

We can then change the current source back to a voltage source. V = (4 − 12j)(1.8 + j2.4) = 36 − 12jV We now have this circuit.

To find the voltage, we can do one last source transformation OR we can use the principle of voltage division to find V0 OR we can find the current going through all components using Ohm’s Law and then multiplying this current only by the impedance represented by the resistor in series with the capacitor. Here, we will find the current then multiply it with the impedance. I0 =

36 − j 12 12(3 − j ) 39 + j 27 = 36.12 − j18.84A = = 25 12 − j16 4(3 − j4)

To find V0 , we multiply the impedance with the current. V0 = I0 (10 − 19j ) = 36.12 − j 18.84V

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Th´ evenin and Norton Equivalent Circuits

Example:

Find the Th´evenin equivalent with respect to terminals a,b in this circuit. Answers: We will first find the Th´evenin voltage. We can first do a source transformation on the independent voltage source in series with the resistor and group the resulting parallel resistors into one equivalent resistance.

Finally, do another source transformation. Note that there will be an empty spot where you previously had Vx . Don’t forget to mark this voltage on your circuits! This is how the circuit will look now.

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To find the current that is marked as I on the diagram, we can write a KVL for the first loop. 100 = 10I − j40I + 120I + 10Vx = (130 − j 40)I + 10Vx But also note that Vx = 100 − 10I (the voltage of the source minus the voltage drop on the resistor in series with it). We can rewrite our previous equation. 100 = (130 − j40)I + 1000 − 100I ⇒ −900 = (30 − 40j)I We brought our equation to the form V = ZI. We can find the current to be I=

−900 = 18∠(−126.87◦ ) 30 − 40j

Since we found the current, we can also find the voltage Vx . Vx = 100 − 180∠(−126.87◦ ) = 208 + j144V To find the VT h , we write a KVL for the second loop (the open loop). −VT h + 10Vx + 120I = 0 VT h = 2080 + j 1440 + j 120 · 18∠(−126.87◦ ) = 835.22∠(−20.17◦ )V

To find the Th´evenin impedance ZT h with respect to terminals a,b, we will use the test voltage source method (since there is a dependent voltage source in the circuit). We replace the terminals a,b with a voltage source VT and we replace all independent sources in the circuit with short circuits if they’re voltage sources or open circuits if they’re current sources. We will get the following equivalent circuit.

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We can group the 2 parallel resistors together and replace them with a 10Ω resistor. The current IT will be split into 2 currents, Ia and Ib . We will anticipate the next topic by claiming that Ia =

VT − 0 10 − j 40

(This is essentially a node voltage current) Since we found Ia, we can find Vx = (12||60)Ia = 10Ia We will find Ib in the same manner (by writing a node voltage equation). Ib =

Vt − 10Vx −VT (9 + j 4) = 120(1 − j 4) 120

We also know that IT = Ia + Ib . IT =

VT − 0 −VT (9 + j 4) VT (3 − 4j ) + = 12(10 − 40j ) 10 − j40 120(1 − j4)

The Th´evenin impedance is the ratio ZT h =

VT IT

,which can be found from the above equation.

VT = 91.2 − j38.4Ω IT

Now that we have both ZT h and VT h , we can draw the equivalent circuit between points a and b.

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Bibliography and further reading All the figures in this presentation have been taken from the book ”Electric Circuits 9th edition” by Nillson and Riedel. For further reading, check pages 317-332 for phasor domain circuit analysis methods. Finally, there are 2 lectures left to be posted online. If you find any errors of any kind (grammar, maths, scripting,typos etc.) , please send an e-mail with a screen shot and a description of the error to [email protected] I am going to rewrite the erroneous parts of all documents and resend them after 15 of April or so. I am also going to delete the e-mail address from all the documents, so please aim to send any remarks by that time ! Thank you for reading this part. Do read the book.

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