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Basic Electro-Mechanical Engineering (EE-170) : Lecture#05

The document is a lecture on basic electro-mechanical engineering that discusses circuit analysis using series and parallel combinations. It provides an example circuit and shows how to solve for currents and voltages by combining resistors that are in series or parallel. Specifically, it combines a 6 and 4 ohm resistor in series to 10 ohms, combines an 8 and 10 ohm resistor in parallel to 14.4 ohms, and combines a 10 and 4.4 ohm resistor in series to 14.4 ohms. It then uses Kirchhoff's laws to solve for the currents and voltages throughout the circuit.

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Hassnain Massid
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67 views50 pages

Basic Electro-Mechanical Engineering (EE-170) : Lecture#05

The document is a lecture on basic electro-mechanical engineering that discusses circuit analysis using series and parallel combinations. It provides an example circuit and shows how to solve for currents and voltages by combining resistors that are in series or parallel. Specifically, it combines a 6 and 4 ohm resistor in series to 10 ohms, combines an 8 and 10 ohm resistor in parallel to 14.4 ohms, and combines a 10 and 4.4 ohm resistor in series to 14.4 ohms. It then uses Kirchhoff's laws to solve for the currents and voltages throughout the circuit.

Uploaded by

Hassnain Massid
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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BASIC ELECTRO-MECHANICAL ENGINEERING

(EE-170)

Lecture#05

Engr. Ayla Safdar


Lecturer
Electrical Department
Example Circuit

Solve for the currents through each resistor


And the voltages across each resistor using
Series and parallel simplification.
Example Circuit

The 6 and 4 ohm resistors are in series, so


are combined into 6+4 = 10Ω
Example Circuit

The 8 and 10 ohm resistors are in parallel, so


are combined into 8∙10/(8+10) =14.4 Ω
Example Circuit

The 10 and 4.4 ohm resistors are in series, so


are combined into 10+4.4 = 14.4Ω
Example Circuit

+
I1∙14.4Ω
-

Writing KVL, I1∙14.4Ω – 50 v = 0


Or I1 = 50 v / 14.4Ω = 3.46 A
Example Circuit
+34.6 v -

+
15.4 v
-

If I1 = 3.46 A, then I1∙10 Ω = 34.6 v


So the voltage across the 8 Ω = 15.4 v
Example Circuit
+ 34.6 v -

+
15.4 v
-

If I2∙8 Ω = 15.4 v, then I2 = 15.4/8 = 1.93 A


By KCL, I1-I2-I3=0, so I3 = I1–I2 = 1.53 A
Three Phase Circuit
Objectives

 Explain the differences between single-phase, two-


phase and three-phase.
 Compute and define the Balanced Three-Phase
voltages.
 Determine the phase and line voltages/currents for
Three-Phase systems.
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Wye and Delta Connections

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Delta Connection

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Delta Connected System

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Delta Connected System

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