Chapter 1 - EE.pptx

8/10/2019 Chapter 1 - EE.pptx

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Chapter 1:

Fundamental of Electric Circuit


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Content

Distinguish the principal elements of electric

circuit: nodes, loops, meshes, branches,

voltage and current source, power and energy.

Apply Kirchhoffs laws to simple electric

circuits and derive the basic circuit equations.

State and Apply Ohms law to electric circuits


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1.1. Distinguish the principal elements

of electric circuit: nodes, loops,

meshes, branches, voltage and currentsource, power and energy.


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An electric circuit is an interconnection of electrical elements.

A simple electric circuit is shown in Fig. 1.1. It consists of three

basic components: a battery, a lamp, and connecting wires.

Such a simple circuit can exist by itself; it has several

applications, such as a torch light, a search light, and so forth.


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Branch

A branch is any portion of a circuit with two

terminals connected to it.


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Node


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Loop

A loop is any closed connection of branches.


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Mesh

A mesh is a loop that does not contain other loops.


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1.2 Apply Kirchhoffs laws to simple

electric circuits and derive the basic

circuit equations.


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KIRCHHOFFS CURRENT

LAW (KCL)

Choose

Add the current entering the

node (and subtract the oneleaving the node).

5 + i - 2 - (-3) = 0

i = -6 A

EnterExit


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EXAMPLE

From Ohms Law,

(eq. 1)iv 21 iv 32

Applying KVL around the loop gives,

(eq. 2)02021

vv


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EXAMPLE (contd)

Substituting (eq. 1) into (eq. 2), we obtain:

Substituting i in (eq. 1) finally gives:

205 i

03220 ii

A4i

V81v V12

2 v


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Given:

Find: v0and i in the circuit shown above.

Solution:

EXAMPLE


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EXAMPLE (contd)

We apply KVL around the loop as shown in the figure.The result is:

(eq. 1)

Applying Ohms Law to the 6- resistor gives:

(eq. 2)

Substituting (eq. 2) into (eq. 1) yields:

04241200

vvi

iv 60

0121016 ii

A8i

V480v


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1.3 State and Apply Ohms law to

electric circuits


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Power is the time rate of expending or absorbing energy,

measured in watts (W).

The electric power generated by an active element, or that

dissipated or stored by a passive element, is equal to the

product of the voltage across the element and the current

flowing through it.

P = V.I

ELECTRIC POWER


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ELECTRIC POWER

In general:

Energy absorbed or supplied by an element from timet0 totime tis:

Energy is the capacity to do work, measured in joules (J).

Power absorbed = - Power supplied

t

t

t

t

dtvidtpw

00


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SIGN CONVENTION

Passive sign convention states that:

The power dissipated by a load is a positive quantity

Power dissipated (+)

Power generated (-)

Power dissipated (-)

Power generated (+)

i

i


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Steps Required:

Choose an arbitrary direction of current flow

Label polarities of all active elements (voltage

and current sources)

Assign polarities to all passive elements

(resistors and load), for passive elements the

current always flow into the positive terminal

Compute the power dissipated by each

element according to the following rule


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Rules:

If positive current flows into positive terminal of

an element, then the power dissipated is positive (

element absorbs power) If the current leaves the positive terminal of an

element, then the power dissipated is negative

(element delivers power)


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Given:

Find: For the circuit of figure shown, determine whichcomponents are absorbing power and which are

delivering power. Is conservation of power

satisfied? Explain your answer.

EXAMPLE


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Solution:

Known Quantities:Currents through elements D and E;

voltage across elements B, C, E.

Find:Which components are absorbing power, which

are supplying power; verify the conservation of power.

Analysis:

Step 1: Choose an arbitrary direction of current flow.

- Current flow is in clockwise direction

Step 2: Label polarities for voltage or current sources

EXAMPLE


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Solution (cont)Therefore

- By KCL, the current through element B is 5 A, to the right.

- By KVL: -va3 +10 + 5 = 0

Therefore, the voltage across element A is:

va= 12 V (positive at the top)

EXAMPLE


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Step 3: Compute the power dissipated by each element

A supplies (12 V)(5 A) = 60 W

B supplies (3 V)(5 A) = 15 W

C absorbs (5 V)(5 A) = 25 W

D absorbs (10 V)(3 A) = 30 W

E absorbs (10 V)(2 A) = 20 W

Total power supplied = 60 W + 15 W = 75 W

Total power absorbed = 25 W+30W+20W= 75 W

Total power supplied = Total power absorbed, so conservation of

power is satisfied

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Chapter 1 - EE.pptx