Part 6 – Combined Circuits

Resistive, Inductive, and Capacitive Reactance

Resistive, Inductive and Capacitive Reactance

Resistance/Inductance Circuit

The circuit consists of resistance (R ohms) in series with inductance (L henrys).

Resistance and Inductance in Series highlighting I, e, L and R
Resistance and Inductance in Series

R is in phase with the current. The voltage applied to an inductance leads 90° in front of the current.

Activity: Voltage and Current Curves

Click the image hotspots to learn more about voltage and current curves.

SLIDE 6 – 12 – make a video explaining the contents and the graphics

Representation of an Alternating Quantity by a Rotating Vector

Capacitance

  • XC = 1/2πfC
  • XC = Capacitive reactance (Ω)
  • f = Frequency (Hz)
  • C = Capacitance (F)

Inductance

  • XL = 2πfL
  • XL = Inductive reactance (Ω)
  • f = Frequency (Hz)
  • L = Inductance (H)

Instantaneous Current

Vectors Representing Current and Voltage Differing in Phase

Addition of Alternating Voltages of Current Using Vectors

The addition of alternating voltages of current using vectors is a common practice in AC analysis. However, it’s important to note that the vector sum of OC is typically less than the arithmetical sum of OA and OB, unless OA and OB are in phase with each other. This phenomenon underscores why it is rarely accurate in AC work to simply add voltages or currents together arithmetically.

Application of Vectors to a Circuit Consisting of Resistance and Inductance in Series

Activity: Application of Vectors to a Circuit Consisting of Resistance and Inductance in Series

Click the ticks on the slider bar to learn more about the application of vectors to a circuit consisting of resistance and inductance in series.

Calculation of Impedance – Resistance and Inductance in Series

Since the angle between OB and BE is a right angle:
OE2 = OB2 + BE2 = OB2 + OC2

Em2 = (R Im)2 + (2πfL Im)2

(R Im)2 = OB = Voltage across resistance

(2πfL Im)2 = OC = Applied voltage

Em = Im√(R2 + (2πfL)2)
= Im√(R2 + XL2)

And Em/Im = √(R2 + XL2)

If E and I are the effective (rms) values of the voltage and the current respectively, then:

    \begin{align*} \frac{E}{I} &= \frac{0.707E_m}{0.707I_m} \\ &= \sqrt{R^2 + X_L^2} \\ &= Z \text{ ohms (Impedance in circuit)} \end{align*}

Where

  • R = Circuit resistance (ohms)
  • XL = Circuit inductive reactance (ohms)
  • L = Inductance (Henrys)

The impedance of any AC circuit is always given by the ratio of the voltage to the current and is represented by the symbol Z.

Z ohms = √(R2 + XL2)

Z ohms = E/I

I amperes = E volts / Z ohms

Triangle highlighting resistance, impedance and inductive reactance

Activity: Vector Diagrams

Click the tick marks to learn more about Vector Diagrams.

When a circuit is non-inductive, its impedance is the same as the resistance. When the circuit consists of an inductor, the impedance is practically the same as the reactance.

\begin{aligned} \cos \Phi & =\frac{\mathrm{OB}}{\mathrm{OE}}=\frac{\mathrm{R} \mathrm{I}_{\mathrm{m}}}{\mathrm{Z} \mathrm{I}_{\mathrm{m}}}=\frac{\mathrm{R}}{\mathrm{Z}} \\ & =\frac{\text { Resistance }}{\text { Impedance }} \end{aligned}

Also

\begin{aligned} \tan \Phi & = \frac{E B}{O B} = \frac{2 \pi f L \mathrm{I}_{\mathrm{m}}}{R \mathrm{I}_{\mathrm{m}}} \\ & =\frac{2 \pi f L}{R} \\ & = \frac{\text{reactance}}{\text{resistance}} \end{aligned}

And

$\sin \Phi=\frac{\text { resistance }}{\text { impedance }}$

Capacitive/Reactance Circuit

\text{Capacitive Reactance (} X_c \text{)} = \frac{1}{2\pi f C}

Capacitive reactance is inversely proportional to capacitance and frequency.

Inductive Reactance (XL) = 2πfL

Inductive reactance is directly proportional to inductance and frequency.

Impedance in a Capacitive/Reactance Circuit

Example:
120 volts AC is connected across a 12 ohm resistor and a 100 μF capacitor.

Solution:
F = 60 Hz
C = 100 μF = 100 x 10-6 F

Capacitive reactance:
XC (ohms)

= 1/(2πfC)

= 1/(2 x 3.1416 x 60 x 100×10-6)

= 2.65 ohms

Impedance in a Capacitive/Reactance Circuit

Example:

120 volts AC is connected across a 12 ohm resistor and a 100 μF capacitor.

the words impedance, resistance and capacitive reactance forming a triangle

Solution:

Impedance is the vector sum of capacitive reactance and resistance:

Z=√(R2 + XC2)

=√(122 + 2.562)

=12.3 ohms

Current is the voltage divided by impedance

I=E/Z

=120/12.3

=9.76 amps

Impedance in a Capacitive/Inductive/Reactance Circuit

Impedance is the vector sum of capacitive reactance, Inductive reactance, and resistance:

XT=(XL – XC)

Z = √(R2 + XT2)

Impedance in a capacitive/inductive/reactance circuit highlighting AC SOURCE R = 120 Xc= -120 (A) Circuit diagram. XL = = 30 Xc = 120 XT=90 IMPEDANCE Z = 150 REACTANCE RESISTANCE R = 120 XL: -30 (B) Vector diagram. RESISTANCE XT = XC-XL IMPEDANCE INDUCTIVE OR REACTANCE IMPEDANCE 120-30 CAPACITIVE REACTANCE -90 RESISTANCE (C) Impedance triangles. Z = √R² +X+2 =√122 +92 = √144 + 81 = √225 = 150 (D) Problem solution. Figure 2-17. Impedance in series AC circuits: (A) Circuit diagram; (B) Vector dia- gram; (C) Impedance triangles, {DJ problem solution

Example:

An AC circuit contains a resistance of 20 ohms connected in series with an induction coil of 80 mH and a 16 μF condenser (capacitor).

Find the total impedance when working on a 60 Hz power supply

Solution:

R = 20 ohms
L = 80 mH = 0.8 H
C = 16 μF = 16 x10-6 F

XT = (XL – XC)
XL = 2 π f L ohms
= 2(3.1416)(60 Hz)(0.8 H)
= 301.6 ohms

XC = 1/2 π f C
XC = 1/2(3.1416)(60 Hz)(16 x 10-6 F)
= 33.2 ohms

Solution:

R = 20 ohms
L = 80 mH
C = 16 μF
XL = 301.6 ohms
XC = 33.2 ohms

XT = (XL – XC)
= 301.6 – 33.2 ohms
= 268.4 ohms

Z = √(R2 + XT2)
= √(202 + 268.42)
= 170 ohms

 

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PEG-3722 Electrotechnology Copyright © by Josee Beaulieu. All Rights Reserved.

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