1. ECE 3301 General Electrical Engineering
Chapter 5.3
Complex Impedances

2. Phasor Circuit Elements
Each of the circuit elements has a phasor domain representation.
This representation incorporates the characteristics of resistance, inductance and capacitance.

3. Resistance
Consider the resistance driven by a sinusoidal voltage source as shown below. 𝑣 𝑖 𝑅
The voltage source is described by the equation.
𝑣= 𝑉 𝑚 cos 𝜔𝑡+𝜙

4. Resistance Applying Ohm’s Law reveals: 𝑣= 𝑉 𝑚 cos 𝜔𝑡+𝜙𝑖 𝑅
𝑣= 𝑉 𝑚 cos 𝜔𝑡+𝜙
𝑖= 𝑣 𝑅 = 𝑉 𝑚 𝑅 cos 𝜔𝑡+𝜙

5. The voltage and the current are “in phase”.
𝑣= 𝑉 𝑚 cos 𝜔𝑡+𝜙
𝑖= 𝑉 𝑚 𝑅 cos 𝜔𝑡+𝜙

6. Resistance The voltage source may be expressed in Phasor Form :
𝑣= 𝑉 𝑚 cos 𝜔𝑡+𝜙
𝑽 = 𝑉 𝑚 ∠𝜙
𝑖= 𝑉 𝑚 𝑅 cos 𝜔𝑡+𝜙
𝑰 = 𝑉 𝑚 𝑅 ∠𝜙

7. Resistance Im
𝑽 = 𝑉 𝑚 ∠𝜙
𝑰 = 𝑉 𝑚 𝑅 ∠𝜙 𝜙 Re

8. Resistance Define the complex impedance of the resistance:
𝒁 𝑅 = 𝑽 𝑰 = 𝑉 𝑚 ∠𝜙 𝑉 𝑚 𝑅 ∠𝜙
𝒁 𝑅 =𝑅

9. Resistance
The complex impedance of the resistance may be graphed in Phasor form: Im
𝒁 𝑅 =𝑅 Re

10. Ohm’s Law The Phasor Domain version of Ohm’s Law is: 𝑽 = 𝑰 𝒁
For a resistance this is:
𝑽 = 𝑰 𝑅

11. Inductance
Consider the inductance driven by a sinusoidal current source as shown below. 𝑣 𝑖 𝐿
The current source is described by the equation.
𝑖= 𝐼 𝑚 cos 𝜔𝑡+𝜙

12. Inductance The defining characteristic of inductance: 𝑣=𝐿 𝑑𝑖 𝑑𝑡
𝑖= 𝐼 𝑚 cos 𝜔𝑡+𝜙 𝐿
𝑣=𝐿 𝑑𝑖 𝑑𝑡
𝑣=− 𝐼 𝑚 𝜔𝐿 sin 𝜔𝑡+𝜙
𝑣=− 𝐼 𝑚 𝜔𝐿 cos 𝜔𝑡+𝜙− 90 °

13. The voltage and the current are 90 degrees out of phase.
90°
The voltage and the current are 90 degrees out of phase.
𝑖= 𝐼 𝑚 cos 𝜔𝑡+𝜙
𝑣=− 𝐼 𝑚 𝜔𝐿 cos 𝜔𝑡+𝜙− 90 °

14. Inductance The voltage and current may be expressed in Phasor Form :
𝑖= 𝐼 𝑚 cos 𝜔𝑡+𝜙
𝑰 = 𝐼 𝑚 ∠𝜙
𝑣=− 𝐼 𝑚 𝜔𝐿 cos 𝜔𝑡+𝜙− 90 °
𝑽 =− 𝐼 𝑚 𝜔𝐿 exp 𝑗 𝜙− 90 °
𝑽 =− 𝐼 𝑚 exp −𝑗 90 ° 𝜔𝐿 exp 𝑗𝜙

15. Inductance Recall that: exp 𝑗 −90° =−𝑗 𝑽 =− 𝐼 𝑚 exp −𝑗 90 ° 𝜔𝐿 exp 𝑗𝜙
𝑽 = 𝐼 𝑚 𝑗𝜔𝐿 ∠𝜙

16. Inductance The Phasors are: 𝑰= 𝐼 𝑚 ∠𝜙 𝑽= 𝐼 𝑚 𝑗𝜔𝐿 ∠𝜙 𝑽= 𝐼 𝑚 ∠𝜙 𝜔𝐿 ∠90°
𝑰= 𝐼 𝑚 ∠𝜙
𝑽= 𝐼 𝑚 𝑗𝜔𝐿 ∠𝜙
𝑽= 𝐼 𝑚 ∠𝜙 𝜔𝐿 ∠90°
𝑽= 𝐼 𝑚 𝜔𝐿 ∠(𝜙+90°)

17. Inductance Im
𝑽 =𝜔𝐿 𝐼 𝑚 ∠ 𝜙+ 90 °
𝑰 = 𝐼 𝑚 ∠𝜙
90 ° 𝜙 Re

18. Inductance Define the complex impedance of the inductance:
𝒁 𝐿 = 𝑽 𝑰 = 𝑗𝜔𝐿 𝐼 𝑚 ∠𝜙 𝐼 𝑚 ∠𝜙
𝒁 𝐿 =𝑗𝜔𝐿
𝒁 𝐿 =𝜔𝐿∠ 90 °

19. Inductance
The complex impedance of the inductance may be graphed in Phasor form: Im
𝒁 𝐿 =𝑗𝜔𝐿=𝜔𝐿∠ 90 ° Re

20. Inductance Note that for 𝜔→0, 𝒁 𝐿 →0, and for 𝜔→∞, 𝑍 𝐿 →∞.
An inductance is a short-circuit at DC and a high impedance at very high frequency.
𝒁 𝐿 =𝑗𝜔𝐿
𝒁 𝐿 =𝜔𝐿∠ 90 °

21. Inductance
Define the Inductive Reactance.
𝑋 𝐿 =𝜔𝐿
𝒁 𝐿 =𝑗 𝑋 𝐿

22. Ohm’s Law The Phasor Domain version of Ohm’s Law is: 𝑽 = 𝐼 𝒁
For an inductance this is:
𝑽 = 𝐼 𝑗𝜔𝐿

23. Example – Time Domain Circuit
𝑣=10 cos⁡(100𝑡+30°) 𝑖 𝑣
𝐿=10 H
Transform this circuit into Phasors:

24. Example – Phasor Domain Circuit
𝑽 =10 ∠ 30° 𝑰 𝑽
𝒁 𝐿 =𝑗𝜔𝐿=𝑗 =𝑗 1000 Ω
𝑰 = 𝑽 𝒁 𝐿 = 10 ∠ 30° 1000 ∠ 90° =0.01 ∠−60°
𝑖=0.01 cos⁡(100𝑡−60°)

25. Phasor Diagram Im 𝒁 𝐿 =1000 ∠ 90° 𝑽 =10 ∠ 30° (not to scale) Re
Current lags Voltage
Characteristic of Inductive Circuits Re
𝑰 =0.01 ∠ −60°

26. 𝑣=10 cos⁡(100𝑡+30°)
𝑖=0.01 cos⁡(100𝑡−60°)

27. Capacitance
Consider the capacitance driven by a sinusoidal voltage source as shown below. 𝑣 𝑖 𝐶
The voltage source is described by the equation.
𝑣= 𝑉 𝑚 cos⁡(𝜔𝑡+𝜙)

28. Capacitance The defining characteristic of capacitance:
𝑣= 𝑉 𝑚 cos⁡(𝜔𝑡+𝜙) 𝑖 𝐶
𝑖=𝐶 𝑑𝑣 𝑑𝑡
𝑖=− 𝑉 𝑚 𝜔𝐶 sin⁡(𝜔𝑡+𝜙)
𝑖=− 𝑉 𝑚 𝜔𝐶 cos⁡(𝜔𝑡+𝜙−90°)

29. The voltage and the current are 90 degrees out of phase.
𝑣= 𝑉 𝑚 cos⁡(𝜔𝑡+𝜙)
𝑖=− 𝑉 𝑚 𝜔𝐶 cos⁡(𝜔𝑡+𝜙−90°)

30. Capacitance The voltage and current may be expressed in Phasor Form :
𝑣= 𝑉 𝑚 cos⁡(𝜔𝑡+𝜙)
𝑽 = 𝑉 𝑚 ∠ 𝜙
𝑖=− 𝑉 𝑚 𝜔𝐶 cos⁡(𝜔𝑡+𝜙−90°)
𝑰 =− 𝑉 𝑚 𝜔𝐶 exp 𝑗⁡(𝜙−90°)
𝑰 =− 𝑉 𝑚 exp − 𝑗 90° 𝜔𝐶 exp⁡( 𝑗𝜙)

31. Capacitance Recall that: exp 𝑗 −90° =−𝑗
𝑰 =− 𝑉 𝑚 exp − 𝑗 90° 𝜔𝐶 exp⁡( 𝑗𝜙)
𝑰 =− 𝑉 𝑚 −𝑗 𝜔𝐶 exp⁡( 𝑗𝜙)
𝑰 = 𝑉 𝑚 𝑗𝜔𝐶 exp⁡( 𝑗𝜙)
𝑰 = 𝑉 𝑚 𝑗𝜔𝐶 ∠𝜙

32. Capacitance The Phasors are: 𝑽 = 𝑉 𝑚 ∠ 𝜙 𝑰 = 𝑉 𝑚 𝑗𝜔𝐶 ∠𝜙
𝑽 = 𝑉 𝑚 ∠ 𝜙
𝑰 = 𝑉 𝑚 𝑗𝜔𝐶 ∠𝜙
𝑰 = 𝑉 𝑚 ∠𝜙 𝜔𝐶 ∠90°
𝑰 = 𝑉 𝑚 𝜔𝐶 ∠ (𝜙+90°)

33. Capacitance Im
𝑽 = 𝑉 𝑚 ∠ 𝜙
𝑰 = 𝑉 𝑚 𝜔𝐶 ∠ 𝜙+ 90 °
90 ° 𝜙 Re

34. Capacitance Define the complex impedance of the capacitance:
𝒁 𝐶 = 𝑽 𝑰 = 𝑉 𝑚 ∠ 𝜙 𝑉 𝑚 𝑗𝜔𝐶 ∠ 𝜙
𝒁 𝐶 = 1 𝑗𝜔𝐶
𝒁 𝐶 = −𝑗 𝜔𝐶 = 1 𝜔𝐶 ∠−90°

35. Capacitance
The complex impedance of the capacitance may be graphed in Phasor form: Im Re
𝒁 𝐶 = 1 𝑗𝜔𝐶 = 1 𝜔𝐶 ∠−90°

36. Capacitance Note that for 𝜔→0, 𝒁 𝐶 →∞, and for 𝜔→∞, 𝑍 𝐶 →0.
A capacitance is an open-circuit at DC and a low impedance at very high frequency.
𝒁 𝐶 = 1 𝑗𝜔𝐶
𝒁 𝐶 = −𝑗 𝜔𝐶 = 1 𝜔𝐶 ∠−90°

37. Capacitance
Define the Capacitive Reactance.
𝑋 𝐶 =− 1 𝜔𝐶
𝒁 𝐶 =𝑗 𝑋 𝐶

38. Ohm’s Law The Phasor Domain version of Ohm’s Law is: 𝑽 = 𝑰 𝒁
For a capacitance this is:
𝑽 = 𝑰 1 𝑗𝜔𝐶

39. Example – Time Domain Circuit
𝑣=10 cos⁡(10𝑡+30°) 𝑖
𝐶=0.001 F
Transform this circuit into Phasors:

40. Example – Phasor Domain Circuit
𝑽 =10 ∠ 30° 𝑖
𝒁 𝐶 = 1 𝑗𝜔𝐶 = 1 𝑗(10)(0.001) =−𝑗100 Ω
𝐼 = 𝑉 𝑍 𝐶 = 10 ∠ 30° 100 ∠=−90° =0.1 ∠ 120°
𝑖=0.1 cos⁡(10𝑡+120°)

41. Phasor Diagram Im 𝑰 =0.1 ∠ 120° 𝑽 =10 ∠ 30° 90° (not to scale) 30°
Current leads Voltage
Characteristic of Capacitive Circuits Re
𝒁=100 ∠ −90°

42. 𝑣=10 cos⁡(10𝑡+30°)
𝑖=0.1 cos⁡(10𝑡+120°)