1. IT2001PA
Engineering Essentials (1/2)
Chapter 13 – Impedance
Lecturer Name
lecturer_email@ite.edu.sg
Sep 4, 2012
Contact Number

2. Chapter 13 – Impedance
Lesson Objectives
Upon completion of this topic, you should be able to:
 Explain and verify the characteristics of a RL and RC
series circuit.
 Calculate the voltage, current and impedance in an AC
circuit.
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3. Chapter 13 – Impedance
Resistive-Inductive Series Circuit
Characteristics of A.C. RL Series Circuit
(1) Current is in phase with VR.
(2) Current lags VL by 90o.
(3) Current lags VS by ϕ where ϕ is the phase angle or
phase difference.
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4. Chapter 13 – Impedance
Resistive-Inductive Series Circuit
In a resistive-capacitive circuit, the opposition to the current
flow is called the impedance.
Symbol : Z
Unit : Ohms ( Ω )
VR = IR XL: Inductive Reactance
VL = IXL
VS = IZ
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5. Chapter 13 – Impedance
Resistive-Inductive Series Circuit
Impedance Triangle of A.C. RL Series Circuit
Impedance Triangle
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6. Chapter 13 – Impedance
Resistive-Capacitive Series Circuit
Characteristics of A.C. RC Series Circuit
(1) Current is in phase with VR.
(2) Current leads VC by 90o.
(3) Current leads VS by ϕ where ϕ is the phase angle or
phase difference.
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7. Chapter 13 – Impedance
Resistive-Capacitive Series Circuit
In a resistive-capacitive circuit, the opposition to the current
flow is called the impedance.
Symbol : Z
Unit : Ohms ( Ω )
VR = IR Xc: Capacitive Reactance
VC = IXC
VS = IZ
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8. Chapter 13 – Impedance
Resistive-Capacitive Series Circuit
Impedance Triangle of A.C. RC Series Circuit
Impedance Triangle
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9. Chapter 13 – Impedance
Quiz on RL Series Circuits
1. An inductive circuit has a resistance of 40 ohms and
an inductance of 0.2 H. It is connected to a 500 V, 50
Hz supply. Find the reactance; impedance and the
current taken.
(Ans: 62.8 ohms ; 74.47 ohms ; 6.71 A)
XL = 2πfL Z = √ (XL2+ R2) I = V/Z
= 2x3.14x50x0.2 = √62.82 + 402 = 500/74.47
= 62.8 Ω = 74.47 Ω = 6.71 A
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10. Chapter 13 – Impedance
Quiz on RL Series Circuits
2. An inductive circuit of resistance 16.5 ohms and
inductance 0.14 H takes a current of 25 A. If the
frequency is 50 Hz, find the impedance and the
supply voltage.
(Ans: 47 ohms ; 1175 V)
XL = 2πfL Z = √ (XL2+ R2) V = IZ
= 2x3.14x50x0.14 = √43.982 + 16.52 = 25x47
= 43.98 Ω = 46.98 Ω = 1175 A
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11. Chapter 13 – Impedance
Quiz on RL Series Circuits
3. A coil of inductance 0.45 H takes a current of 0.85 A
when connected to 250 V, 40 Hz supply. What is the
resistance of the coil?
(Ans: 271.5 ohms)
XL = 2πfL Z = V/I Z2 = (XL2+ R2)
= 2x3.14x40x0.45 = 250/0.85 294.12 = 113.12 + R2
= 113.1 Ω = 294.1 Ω R = 271.5 Ω
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12. Chapter 13 – Impedance
Quiz on RL Series Circuits
4. An inductive circuit of resistance 4.6 ohms takes a
current of 1.2 A when connected to a 220 V, 50 Hz
supply. What is the inductance of the coil?
(Ans: 0.583 H)
Z = V/I Z2 = (XL2+ R2) XL = 2πfL
= 220/1.2 183.32 = XL2 + 4.62 L = XL/2πf
= 183.3 Ω XL = 183.24 Ω = 183.24/2x3.14x50
L = 0.583 H
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13. Chapter 13 – Impedance
Quiz on RL Series Circuits
5. A coil of inductance 0.02 H is connected in series with
a resistance of 5 ohms across a supply of 200 V, 50
Hz. Find the circuit current, voltage across the
resistor and inductor and the phase angle.
(Ans: 24.91 A ; 124.55 V ; 156.5 V ; 51.5°)
XL = 2πfL Z2 = (XL2+ R2) I = V/Z
= 2x3.14x50x0.02 Z2 = 6.282 + 52 = 200/8.03
XL = 6.28 H Z = 8.03 Ω = 24.91 A
VR = IR VL = IXL tan-1ϕ = XL/R
= 24.91x5 = 24.91x6.28 = 156.5/124.55
= 124.55 V = 156.5 V ϕ = 51.5°
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14. Chapter 13 – Impedance
Quiz on RC Series Circuits
1. A capacitive reactance of 40 ohms is connected in
series with a resistor of 20 ohms across a 110 V
supply. Find the circuit current.
(Ans: 2.46 A)
Z2 = (XL2+ R2) I = V/Z
Z2 = 402 + 202 = 110/44.72
Z = 44.72 Ω I = 2.46 A
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15. Chapter 13 – Impedance
Quiz on RL Series Circuits
2. A capacitor of 50 µF capacitance is connected in
series with a resistor of 12 ohms across a 100 V, 50
Hz supply. Find the impedance, current and voltage
across each component.
(Ans: 64.78 ohms ; 1.54 A ; 18.53 V ; 98.29 V)
Xc = 1/2πfC Z2 = (XL2+ R2)
= 1/(2x3.14x50x50x10-6) Z2 = 63.662 + 122
Xc = 63.66 Ω Z = 64.78 Ω
I = V/Z VR = IR VC = IXc
= 100/64.78 = 1.54 x 12 = 1.54 x 63.66
= 1.54 A = 18.48 V = 98.0 V
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16. Chapter 13 – Impedance
Quiz on RC Series Circuits
3. A capacitive reactance of 50 ohms is connected in
series with a 15 ohms resistor across a 200 V, 50 Hz
supply. Calculate the impedance, current and phase
angle of the circuit. (Ans: 52.2 ohms ; 3.83 A ; 73.3°)
Z2 = (XC2+ R2) I = V/Z tan-1ϕ = XC/R
Z2 = 502 + 152 = 200/52.2
= 50/15
Z = 52.2 Ω I = 3.83 A
ϕ = 73.3°
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17. Chapter 13 – Impedance
Quiz on RC Series Circuits
4. A capacitive reactance of 70 ohms is connected in
series with a resistance of 22 ohms across an ac
supply. If the supply current is 10 A, calculate supply
voltage. (Ans: 733.76 V)
Z2 = (XC2+ R2) V = IZ
Z2 = 702 + 222 = 10x73.376
Z = 73.376 Ω V = 733.76 V
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18. Chapter 13 – Impedance
Quiz on RC Series Circuits
5. A capacitor C is connected in series with a 40 ohms
resistor across a supply of frequency 60 Hz. A
current of 3 A flows and the circuit impedance is 50
ohms. Calculate the value of capacitance C and the
supply voltage. (Ans: 88.42 µ F ; 150 V)
Z2 = (XC2+ R2) V = IZ Xc = 1/2πfC
502 = XC2 + 402 = 3x50 C = 1/2πfXc
XC = 30 Ω V = 150 V = 1/(2x3.14x60x30)
Xc = 88.42µF
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19. Chapter 13 – Impedance
Summary
 Phase shift, phase angle, characteristics of
 Resistive-Capacitive Circuit
 Resistive-Inductive Circuit
 Calculation of voltage, current and impedance in ac
circuits.
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20. Chapter 13 – Impedance
Next Lesson
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