AC CIRCUIT ANALYSIS USING PHASORS AND EQUIVALENT IMPEDANCE CONCEPT Week 9 AC CIRCUIT ANALYSIS USING PHASORS AND EQUIVALENT IMPEDANCE CONCEPT

Learning Outcomes After completing this topic you will be able to: Apply the phasor concept in ac-driven circuit analysis. Apply the phasor i-v relationships of the resistor, capacitor, and inductor in steady-state ac circuit analysis. Apply the equivalent impedance concept rule to find the magnitude and phase relationships of currents and voltages in ac circuits. Draw phasor diagrams for the circuit voltages and currents. 1/1/2019 2 2

Phasors and Circuit Analysis To use phasors for circuit analysis, we first need to know how KVL, KCL, and Ohm’s law in apply in the phasor domain.

Kirchhoff’s Voltage Law in the Phasor Domain Consider the following time-domain circuit where a sinusoidally varying voltage source v1(t) is driving a current i(t) through the circuit and establishing voltage drops v2(t), v3(t).... Vn(t) across the rest of the circuit elements. + v2(t) - + v3(t) - + vn(t) - + v1(t) -

Recall that KVL states that the sum of voltages around a closed path is equal to zero. 0 = v1(t) + v2(t) + …. + vn(t) For a sinusoidally driven circuit, we can write

In general, ejt ≠ 0, so This result tells us that KVL applies in the phasor domain as well as the time domain. + v2(t) - + v3(t) - + vn(t) - Time-domain circuit + v1(t) - Phasor transform + V2 - + V3 - + Vn - + V1 - Phasor-domain circuit 1/1/2019 6

Kirchhoff’s Current Law in the Phasor Domain Recall that KCL states that the sum of currents leaving (or entering) a node is equal to zero. 0 = i1(t) + i2(t) + …. + in(t) i(t) i1(t) i2(t) in(t)

Using Euler’s identity ej = cos + jsin, we can rewrite the previous equation as

In general, ejt ≠ 0, so This result shows that KVL applies in the phasor domain as well as the time domain. Phasor-domain KCL equation Time-domain KCL equation i(t) i1(t) i2(t) iN(t)

Driving Point Impedance of a Two-Terminal Network The driving point impedance of a two-terminal circuit (box) is the impedance seen by a source V connected across its input terminals. The value of the driving point impedance is given by the ratio of the voltage and current phasor: Z Note When Z applies to a 2-terminal circuit (rather than simple component) it is known as the equivalent impedance, or alternatively, as the driving point impedance of the box.

Driving Point Impedance of Series-Connected Impedances Consider the following circuit where voltage source V is supplying current I to a set N impedances connected together in series and lumped in an imaginary box. Suppose we wish to find the equivalent impedance seen by the source V. To do this we need to find the ratio V/I of the box. Z Application of KVL to the circuit leads us to the equation Therefore, Equivalent impedance of box seen by source V is given by the ratio of V/I. Z

Driving Point Impedance of Parallel-Connected Impedances For the circuit shown below, application of KCL leads to the node current equation Now, application of Ohm’s law to the impedances gives us the following relationships Z1 Z2 ZN

Thus, upon substitution of the above expressions into the KCL equation, we obtain We can rewrite the above equation as

Hence, we obtain the result This means we can replace the N parallel-connected impedances with a single impedance, Z. Z ZN Z1 Z2

For the simple case of two impedance connected in series, we obtain This can be rewritten in the more convenient form as Z2 Z1 Z

Worked Example 1 Determine the driving-point impedance of the circuit at a frequency of 20 Hz. = 100μF

Solution

Worked Example 2 Determine the driving-point admittance of the circuit at a frequency of 400 Hz.

Worked Example 3 Determine the input impedance of the circuit at  = 10 rad/sec. Z1 Z3

Solution Let Z1 = impedance of the 2-mF capacitor in series with the 20-Ω resistor Z2 = impedance of the 4-mF capacitor Z3 = impedance of the 2-H inductor in series with the 50-Ω resistor Z1 Z3

Z1 Z3

Exercise Determine the driving-point impedance of the circuit at a frequency of 20 Hz. Z I + V -

Worked Example 4 Determine the driving point impedance of the circuit at a frequency of 40 kHz.

Solution Convert the resistance and capacitance into impedances. ZR = 25 Ω ZC = -j19.89 Ω Z Therefore, driving point impedance of box is

Phasor Analysis Steps Phasor analysis (a.k.a sinusoidal steady-state analysis) consists basically of four steps. Transform all independent sources to phasors Calculate the impedance of the passive circuit elements Apply analysis methods learned in Electric Circuit Analysis I Apply inverse phasor transform to obtain time-domain expression for currents and voltages of interest. Note All of the sources must have the SAME frequency! 25

Worked Example 5 For the circuit shown below, find: Angular frequency (in radians per second) Impedance of R in Ω Impedance of L in Ω Impedance of C in Ω Driving point impedance in Ω Phasor voltage and current Find particular response i(t) 26

Solution Angular frequency  = 1000 rad/s Impedance of R in Ω: ZR = R Ω Impedance of L in Ω: ZL = jL = j x 1000 x 25 x 10-3 = j25 Ω Impedance of C in Ω: ZC = 1/(jC) = 1/(j x 1000 x 10 x 10-6) = -j100 Ω

Z Driving point impedance, Zin = ZR + ZC + ZL = 50 – j100 + j25 = 50 - j75 Ω = 90.14-56.14° Z

Find phasor voltage and current. VS = P{vS(t)} = P{35cos(1000t + 30°)} = 3530° V I = P{i(t)} I Zin VS Application of KVL to the circuit leads us to the equation - VS + IZin = 0 Hence, I = VS Zin 3530° 90.14-56.14° = 0.38886.14° A

Find particular response i(t). i(t) = 0.388cos(1000t + 86.14°) A Phasor diagram: VL VS VR VC I 75° 30° 15°

Worked Example 6 Determine v(t) and i(t).

Solution Draw the phasor equivalent circuit. ; Therefore, VS I j2 Ω

Solution 2. Find equivalent impedance seen by the source. 3. Find circuit current. 4. Find phasor voltage drop V across the inductor. 5. Find time-domain current and voltage.

Worked Example 7 Determine Current i(t) voltage drop vL(t) voltage drop vR(t)

Solution Draw the phasor equivalent circuit. ; ; Therefore, I 250 Ω + VL - + VR - VS j250 Ω

Solution 2. Find equivalent impedance seen by the source. 3. Find circuit current. 4. Find phasor voltage drop VL across the inductor. 5. Find phasor voltage drop VR across the resistor.

Solution 6. Find time-domain current and voltages.

Worked Example 8 Find the steady-state expression for io(t) if ig(t) = 125cos(500t) mA R1 = 50 Ω R2 = 250 Ω L = 1 H C = 20 F R1 R2 C L

Solution Angular frequency  = 500 rad/s Impedance of R1 in Ω: ZR1 = 50 Ω Impedance of R2 in Ω: ZR2 = 250 Ω Impedance of L in Ω: ZL = jL = j x 500 x 1 = j500 Ω Impedance of C in Ω: ZC = 1/(jC) = 1/(j x 500 x 20 x 10-6) = -j100 Ω

Driving point impedance, Zin = (ZR1 + ZC) // (ZR2 + ZL) = (50 – j100)//(250 + j500) = 75 - j100 Ω = 125-36.87 Zin

Find phasor voltage and current. Io = P{io(t)} = P{125cos(500t )} = 1250° mA Vo = P{vo(t)} Application of Ohm’s law to the circuit leads us to the equation Zin Io Vo Vo = IoZin Hence, Vo = IoZin = 125 x 10-3 0 x 125-36.87 = 15.625-53.13 V

Now, VO 15.625-53.13 IG = = = 0.028-116.56 A 250 + j500 ZR2 + ZL Find particular response iG(t). iG(t) = 0.028cos(500t - 116.56°) A

Exercise Find the steady-state expression for vo(t) if vg(t) = 64cos(8000t)

Solution Angular frequency  = 8000 rad/s Impedance of R in Ω: ZR1 = 2kΩ Impedance of L in Ω: ZL = jL = j x 8000 x 500 x 10-3 = j4500 Ω Impedance of C in Ω: ZC = 1/(jC) = 1/(j x 8000 x 31.25 x 10-9) = -j4000 Ω

Driving point impedance, Zin = ZC + ZR // ZL = -j4000 + (2000//j4500) = 1670 – j4742 Ω = 502870.6 ZC ZL ZR ZIN

Find phasor voltage and current. Vg = P{vg(t)} = P{64cos(8000t )} = 640° V I = P{i(t)} I Application of KVL to the circuit leads us to the equation Zin VS - Vg + IZin = 0 Hence, I = Vg Zin 640° 502870.6° = 0.013-70.6° A

Find phasor voltage drop VL across the inductor. Find particular response vO(t). vo(t) = 89.974cos(8000t +19.4°) A

Concept of Reference Phasor When no reference angle is given either for the supply voltage or the supply current, we can define our own reference angle. The following examples show the use of the reference phasor concept in ac circuit analysis.

Worked Example 8 An a.c. circuit consists of a 25  resistor in series with a 40 F capacitor connected across a 240 V 50 Hz supply. Sketch the circuit and calculate the current in the circuit, and its phase angle relative to the supply voltage.

Solution The time-domain circuit referred to by the question is shown in the following figure. Its phasor equivalent is also show. I vS R C vR vC i VR R VS VC ZC (b) (a)

Notes The supply voltage is given as an rms value. Because our definition of phasor uses peak value for the magnitude of the phasor, we need to multiply the rms value with 2 to convert it into peak value. The question does not give any information on the phase angle of the supply voltage. To represent the supply voltage with a phasor, we need to assign any arbitrary angle to it. For simplicity, we choose 0 as the phase angle for the supply voltage phasor. That is, we write By assigning an angle to the supply voltage phasor, it means that we have chosen it to be the reference phasor. The angles of other phasors in the circuit are measured with respect to this reference angle.

The impedance seen by the source is The supply current is Therefore,

The time domain current is The phasor diagram for circuit is shown in Figure 8. Since the supply voltage phasor has a phase angle of 0o and the current phasor has a phase angle of -72.56o, the current phasor lags behind the supply voltage phasor by 72.56o. VS Re Im I

Worked Example 9 A 10 Ω resistor is connected in series with a 120 μF capacitor across a 20 V, 100 Hz supply. Calculate impedance of the circuit, current, and voltage drop across each component. Solution The phasor domain circuit is shown in the figure below. VS R C VR VC I

The total circuit impedance is given by Substituting in the given values for R and C into the above equation, we obtain Therefore,

Let VS be the reference phasor so that we can write The circuit current is given by Therefore,

The phasor voltage drop VR is given by The phasor voltage drop VC is given by

Worked Example 10 A resistor and a capacitor are connected in series across a 240 V, 50 Hz supply. If the current flow in the circuit is 5 A, leading the applied voltage by 45.6o, calculate impedance of the circuit phase angle, the resistance and the capacitance the voltage drop across each component.

The phasor domain circuit is shown in the figure. Solution The phasor domain circuit is shown in the figure. Let VS be the reference phasor so that we can write Given that the phasor current I has a magnitude of 5 A and leads the applied voltage by 45.6o, so we can write VS R C VR VC I Therefore,

Referring to the impedance triangle shown below, we obtain Again referring to the impedance triangle in Figure 10, we obtain giving Z Re Im R XC= 1/ωC φ

Worked Example 11 A circuit having a resistance of 25 Ω and an inductance of 12.5 mH is connected to a 100 V, 400 Hz a.c. supply. Calculate the current in the circuit, and its phase angle relative to the supply voltage. Solution The circuit referred to by the question is shown in the following figure. We note that in this problem we are not given the circuit but have to draw it ourselves and provide our own labels and reference directions for the circuit’s current and voltages. 100 V rms 400 Hz 25 Ω 12.5 mH

Solution To use phasor analysis technique to solve the problem, we first need to label the circuit current and the various voltage drops in the circuit and assign a reference direction for each. The labelled time-domain circuit is shown in the figure. 25 Ω 12.5 mH i vR vL Note that since we have not been given the phase angle of the supply voltage, we have conveniently assumed it to be zero. In other words, we have assumed the supply voltage as the reference phasor.

The phasor domain circuit is shown in the figure below. jXL = j2πx400x12.5 x10-3 = j31.416 Ω 25 Ω jXL = j31.416 Ω I VR VL Note To convert rms value of a voltage or a current to peak value, multiply the given rms value with

Ohm’s law gives Substituting the constraint equations obtained above into the KVL equation we get or,

Therefore, The time domain current is

The phasor diagram for circuit is shown in the following figure The phasor diagram for circuit is shown in the following figure. Since the supply voltage phasor has a phase angle of 0o and the current phasor has a phase angle of -51.49o, the current phasor lags behind the supply voltage phasor by 51.49o. VS Re Im I

Exercise Given the following circuit, determine impedance of the circuit the source current the voltage drop across each component. Sketch the impedance and phasor diagram. 8 V rms 20 kHz 3.3 kΩ 15 mH

Given the following circuit, determine impedance of the circuit Exercise Given the following circuit, determine impedance of the circuit the source current the voltage drop across each component. Sketch the impedance and phasor diagram. 4 kΩ XL = 6 kΩ

Exercise A series circuit consisting of a resistance of 50 Ω, an inductance of 0.4 H, and a capacitance of 100 F is connected to a 240 V, 50 Hz single-phase supply. Calculate (a) the impedance of the circuit (b) The current flowing, and (c) The p.d. across each component (d) The phase angle

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