Complex Number Exercises Questions: 1 2 3

Complex Impedance Questions: 1 2 3 4

PHYS 2420 Complex Impedance Solutions

Complex Number Exercises

1. Verify

(a)

(b)

(c)

The key is to convert the denominators to a purely real number by multiplying the correct complex conjugate. Note that (a+bj)(c-dj) = (ac+bd) + (ad+bc)j and (a+bj)(a-bj) = a² + b².

(a) 

(b) 

(c) 



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2. Convert the following to the form re^jφ :

(a)

(b)

(c)

Note that a+bj = re^jφ implies that r² = a² + b² and φ = arctan(b/a). The angles φ are most correctly given in radians but the use of degrees is common.  Warning! Most calculators will give the wrong result when the angle is in the second and third quadrants.

1. 
2. 
3. 
 

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3. Express the following in the form a+jb:

(a)		(b)		(c)
(d)		(e)		(f)

We use the identity e^jφ = cosφ + jsinφ to convert the following.

1. 
2. 
3. 
4. 
5. 
6. 

Complex Impedance

1. Calculate the equivalent impedance for the circuit in the diagram below at a frequency of 100 Hz. Repeat for 1000 Hz.

 

 
 




We first find the complex impedance of each element.



First we do the case where f = 100 Hz. For capacitors, Z = -j/ΩC. Therefore Z₁ = -j/(2π)(100 Hz)(0.01 μF) = -j(1.59155 × 10⁵Ω) and Z₂ = -j/(2π)(100 Hz)(0.1μF) = -j(1.59155 × 10⁴ Ω). For inductors, Z = jΩL. Therefore Z₃ = j2π(100 Hz)(250 mH) = j(1.57080 × 10² Ω) and Z₄ = j2π(100 Hz)(10 mH) = j(6.28319 Ω).

The impedance of the branch with Z₃ and the resistor is

Z₅ = R + Z₃ = (10 Ω) + j(1.57080 × 10² Ω).

Now the Z₅ impedance branch is in parallel with the Z₂ branch, so 1/Z₆ = 1/Z₅ + 1/Z₂. Using Excel or Maple to handle the tedious calculation

Z₆ = 10.200 + 158.639j .

Finally, Z₆ is in series with Z₁ and Z₄, so the net impedance is

Z_net = Z₁ + Z₆ + Z₄ = 10.200 - 158990.021j .

Finding the impedance when f = 1000 Hz is done the same way. Now Z₁ = -j/(2π)(1000 Hz)(0.01 μF) = -j(1.59155 × 10⁴ Ω) and Z₂ = -j/(2π)(1000 Hz)(0.1μF) = -j(1.59155 × 10³ Ω). As well, Z₃ = j2π(1000 Hz)(250 mH) = j(1.57080 × 10³ Ω) and Z₄ = j2π(1000 Hz)(10 mH) = j(62.8319 Ω).

The impedance of the branch with Z₃ and the resistor is

Z₅ = R + Z₃ = (10 Ω) + j(1.57080 × 10³ Ω).

Now the Z₅ impedance branch is in parallel with the Z₂ branch, so 1/Z₆ = 1/Z₅ + 1/Z₂. Using Excel or Maple to handle the tedious calculation

Z₆ = 47730.751 + 97464.576j Ω .

Finally, Z₆ is in series with Z₁ and Z₄, so the net impedance is

Z_net = Z₁ + Z₆ + Z₄ = 47730.751 + 81611.912j Ω .



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2. Determine the current and RMS voltages across each component in the series LCR circuit shown below at a frequency of 900 Hz. Show that, considering the phase angles of the voltages, Kirchhoff’s rule concerning voltages about a loop is valid.




First we determine the complex impedance of each element.




 
 


For inductors, Z = jΩL. Therefore Z₁ = j2π(900 Hz)(250 mH) = j(1.41372 × 10³ Ω). For capacitors, Z = -j/ΩC. Therefore Z₂ = -j/(2π)(900 Hz)(0.1 μF) = -j(1.75839 × 10³ Ω). The three elements are in series so the net impedance is

Z = Z₁ + Z₂ + R = 100 - 354.672j,

where it is wise to use EXCEL or MAPLE to handle the calculations.

Since we are looking for RMS values, it is wise to convert to exponential notation,

Z = 368.4995e^-j1.29598 .

We can write the emf as ε = (10 V)e^jΩt, where Ω = 2π f. Then the current leaving the emf is given by the complex version of Ohm's Law

I = ε /Z = (0.027137 A) e^{j(Ωt + 1.29598)}.

Note that since the circuit is capacitive, Z₂ > Z₁, the current leads the emf.

Similarly we find the voltage drop over each element

V₁ = IZ₁ = [(0.027137 A) e^{j(Ωt + 1.29598)}] j(1.41372 × 10³ Ω) = (38.364 V) e^{j(Ωt + 2.86678)} ,

V₂ = IZ₂ = [(0.027137 A) e^{j(Ωt + 1.29598)}] j(-1.75839 × 10³ Ω) = (47.989 V) e^{j(Ωt - 0.27482)} ,

and

V_R = IR = [(0.027137 A) e^{j(Ωt + 1.29598)}] (100 Ω) = (2.714 V) e^{j(Ωt + 1.29598)} .

The RMS value of each voltage is V_max/Ö2. So V_1-RMS = 27.128 Volts, V_2-RMS = 33.933 Volts, and V_R-RMS = 1.919 Volts.

These values might lead you to believe that Kirchhoff's voltage rule is invalid. However if we write each voltage drop in the form a + ib, we have

V₁ = -36.9245179720055 + 10.4109045889587j ,

V₂ = 46.1880971153873 - 13.0227799474165j ,

and

V₃ = 0.736420856618163 + 2.61187535845786j .

If we add these terms together we get

V₁ + V₂ + V₃ = 10.0 Volts,

as is required.



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3. Determine the RMS current in the 1000 Ω resistor in the circuit below. Is the circuit inductive or capacitive?








First we determine the complex impedance of each element.




 
 


For inductors, Z = jΩL. Therefore Z₁ = j2π(1000 Hz)(10 mH) = j(62.83185 Ω) and Z₃ = j2π(1000 Hz)(250 mH) = j(1.57080 × 10³ Ω). For capacitors, Z = -j/ΩC. Therefore Z₂ = -j/(2π)(1000 Hz)(0.1 μF) = -j(1.59155 × 10³ Ω).

The Z₃ impedance and the 1000 Ω resistor are in parallel; the equivalent impedance is

1/Z_A = 1/Z₃ + 1/R.

Solving for Z_A yields

Z_A = 711.59956 + 453.018350j Ω .

The elements Z₁, Z₂, and Z_A are all in series so

Z_B = Z₁ + Z₂ + Z_A = 711.59956 - 1075.69923j Ω = (1289.768 Ω)e^{-j(56.51446° )} .

To determine the current, we first note that the emf can be written as V(t) = (10 V)e^jΩt, where Ω = 2π f. Then the current is given by the complex version of Ohm's Law

I(t) = V(t)/Z_B = (10 V)e^jΩt / (1289.768 Ω)e^{-j(56.51446° )} = (0.007753 A)e^{j(Ωt + 56.51446° )} .

If we prefer we could write V(t) and I(t) as

V(t) = (10 V)sin(Ωt),

and

I(t) = (0.007753 A)sin(Ωt + 56.51° ) .

The RMS value of the current is given by

I_RMS = I_max/Ö2 = 5.48 mA .

The current leads the emf, so the circuit is capacitive.



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4. The diagram below represents a real inductor, i.e. one with a small resistance, in parallel with a capacitor. Obtain an expression for the inverse complex impedance, 1/Z, as a function of Ω . Plot 1/|Z| versus Ωto determine the resonance frequency. What is the impedance at resonance?





 
 


The inductor and resistor are in series so their equivalent impedance is Z_A = R + jΩL. The capacitor and Z_A are in parallel, so their equivalent impedance is

1/Z = 1/Z_A + 1/(-j/ΩC) = 1/[R + jΩL] + jΩC .

Using MAPLE to solve the above for Z yields

1/Z = R/(R²+Ω ²L²) + j(ΩC - ΩL/(R²+Ω ²L²)) ,

and its absolute value is

.

The graph looks like



Note the minimum around Ω = 100,000 rad/s.



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Questions? mike.coombes@kwantlen.ca