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• Harry King Corral Avenido

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2. Example 1

Find the combined impedance of the following circuit:

Answer Call the impedance given by the top part of the circuit Z1 and the impedance given by the bottom partZ2. We see that Z1 = 70 + 60j Ω and Z2 = 40 − 25j Ω So

ZT=Z1+Z2Z1Z2 =(70+60j)+(40−25j)(70+60j)(40−25j) =110+35j(70+60j)(40−25j) (Adding complex numbers should be done in rectangular form.) Now, we convert everything to polar form and then multiply and divide as follows:

ZT=110+35j(70+60j)(40−25j) =115.4∠17.65o(92.20∠40.60o)(47.17∠−32.01o) (We do the product on the top first.)

=115.4∠17.65o(92.20×47.17)∠(40.60o−32.01o) =115.4∠17.65o4349.074∠8.59o (Now we do the division.)

=115.44349.074∠(8.59o−17.65o) =37.69∠−9.06o (We convert back to rectangular form.)

=37.22−5.93j (When multiplying complex numbers in polar form, we multiply the r terms (the numbers out the front) and add the angles. When dividing complex numbers in polar form, we divide the r terms and subtract the angles. See the Products and Quotients section for more information.) So we conclude that the combined impedance is

ZT=37−5.9j Ω

3.

Example 2

Given that Z1= 200 − 40j Ω and Z2= 60 + 130j Ω,

find

a) the total impedance b) the phase angle c) the total line current Answer a) ZT=Z1+Z2Z1Z2

=(200−40j)+(60+130j)(200−40j)(60+130j) =260+90j(200−40j)(60+130j) =(275.1∠19.09∘)(204.0∠−11.31∘)(143.2∠65.22∘) =275.1204.0×143.2∠(−11.31∘+65.22∘−19.09∘) =106.2∠34.82∘ =87.18+60.64j So we conclude that the total impedance is

ZT=87.2+60.6j Ω b) We see from the second last line of our last answer that the phase angle is ≈35∘. c) Total line current: We use  V = IZ,  the fact that the impedance is 106.2∠34.82∘ and  the fact that the voltage supplied is 12V=12∠0∘

V.

So

I=ZV =106.2∠34.82∘12∠0∘ =0.113∠−34.82∘A 4. Example 3

A 100 Ω resistor, a 0.0200 H inductor and a 1.20 μF capacitor are connected in parallel with a circuit made up of a 110 Ω resistor in series with a 2.40 μF capacitor. A supply of 150 V, 60 Hz is connected to the circuit. Calculate the total current taken from the supply and its phase angle. Answer

For Z1 (the upper part of the circuit), we have:

X = 2πfL = 2π (60)(0.0200) = 7.540 Ω L

XC=2π(60)(1.20×10−6)1=2210.485 Ω

Z = R + j(X − X ) 1

1

L

C

= 100 + j(7.540 - 2210.485) = 100 − 2202.9j

=2205.21∠−87.40∘ Ω For Z2 (the lower part of the circuit), we have:

XC=2π(60)(2.40×10−6)1=1105.243 Ω Z = R + j(X - X ) 2

2

L

C

= 110 + j(−1105.243)

=1110.7∠−84.32∘ Ω So the total impedance, ZT, is given by:

ZT=Z1+Z2Z1Z2 =210−3308.188j2449326.75∠−171.72o =3314.85∠−86.37o2449326.75∠−171.72o =738.9∠−85.35o This last line in rectangular form is ZT = 59.9 − 736.5j Ω Now:

IT=ZTVT =738.9∠−85.35o150∠0o =0.203∠85.35o

So the total current taken from the supply is of the current is ≈85∘.

203 mA and the phase angle

5. Example 1

Find (1 - 2j)6 Answer First, we express 1−2j in polar form:

√12+(−2)2=√5, and arctan(1−2)=296.6o So

1−2j=√5 ∠ 296.6o Then

(1−2j)6=(√5)6∠ [6×296.6o] =125 ∠ [1779.3903o] =125 ∠ [339.39o] (The last line is true because 360∘×4=1440∘, and we substract this from 1779.39∘.) In rectangular form,

x = 125 cos 339.39° = 117 y = 125 sin 339.39° = -44

So (1 - 2j)6= 117 - 44j

6. Example 3

Find the two square roots of −5+12j. Answer For the first root, we need to find √−5+12j. This is the same as (-5 + 12j)1/2. We express −5 + 12j in polar form:

r=√(−5)2+122=13 For the angle:

α=tan−1(xy)=tan−1(512)≈67.38o The complex number −5 + 12j is in the second quadrant, so

θ = 180° − 67.38 = 112.62° So

−5+12j=13∠112.62∘ Using DeMoivre's Theorem:

(r∠θ)n=(rn∠nθ), we have:

(−5+12j)1/2

=131/2∠(21×112.62∘) =3.61∠56.31∘ This is the first square root. In rectangular form,

x = 3.61 cos56.31° = 2 y = 3.61 sin56.31° = 3 So the first root is 2 + 3j. CHECK: (2 + 3j)2 = 4 + 12j - 9 = -5 + 12j [Checks OK] To obtain the other square root, we apply the fact that if we need to find n roots they will be n360o apart. In this case, n=2, so our roots are 180∘ apart. Adding 180∘ to our first root, we have:

x = 3.61 cos(56.31° + 180°) = 3.61 cos(236.31°) = -2 y = 3.61 sin(56.31° + 180°) = 3.61 sin(236.31°) = -3 So our second root is −2−3j. So the two square roots of −5−12j are 2+3j and −2−3j.

7.

Exercises:

1. Evaluate (2∠135∘)8

Answer

(2∠135∘)8=(2)8∠(135∘×8) =256∠1080∘ =256

8.\ Example 1

Express 5(cos135∘+j

sin 135∘) in exponential form.

Answer We have r=5 from the question. We must express θ=135∘ in radians. Recall:

1o=180π So

135o=180135π =43π ≈2.36 radians So we can write

5(cos 135o+j sin135o)

=5e43πj ≈5e2.36j

9.

Express −1+5j in exponential form. Answer

We need to find θ in radians (see Trigonometric Functions of Any Angle if you need a reminder about reference angles) and r.

α=tan−1(xy)=tan−1(15)≈1.37 radians [This is 78.7∘ if we were working in degrees.] Because our angle is in the second quadrant, we need to apply:

θ=π−1.37≈1.77 And

r=√x2+y2 =√(−1)2+(5)2 =√26 ≈5.10 So −1+5j in exponential form is 5.10e1.77j

10. 3. Express in polar and rectangular forms:

2.50e3.84j

Answer We can immediately write:

2.50e3.84j=2.50 ∠ 3.84 [polar form, θ in radians] OR, if you prefer, since 3.84

radians=220∘,

2.50e3.84j=2.50(cos 220∘+j sin 220∘) [polar form, θ in degrees] And, using this result, we can multiply the right hand side to give:

2.50(cos 220∘+j sin 220∘)=−1.92−1.61j Summary Our complex number can be written in the following equivalent forms:

2.50e3.84j [exponential form] 2.50 ∠ 3.84=2.50(cos 220∘+j sin 220∘) [polar form]

−1.92−1.61j [rectangular form]