ENGG ZC111-L3

BITS Pilani, Pilani Campus

Electrical and Electronics Technology

ENGG ZC111LECTURE -2


KIRCHHOFF'S RULES

• KIRCHHOFF'S RULES are used in conjunction with Ohm's law in solving problems involving complex circuits:

• KIRCHHOFF'S FIRST RULE or JUNCTION RULE or CURRENT LAW: The sum of all currents entering any junction point equals the sum of all currents leaving the junction point. This rule is based on the law of conservation of electric charge.

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“The sum of all currents entering a point is equal zero.”

I1

I4 I3

I2

I1 - I2 - I3 + I4 = 0 or

-I1 + I2 + I3 - I4 = 0

Kirchhoff’s Current Law (KCL)

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KCL Metaphor

From the pipe that is full of water, the amount of flow-in water must be equal to the amount of flow-out water.

This is because water cannot disappear.



KIRCHHOFF'S SECOND RULE or LOOP RULE or VOLTAGE LAW: The algebraic sum of all the gains and losses of potential around any closed path must equal zero. This law is based on the law of conservation of energy.

Kirchhoff’s Voltage Law (KVL)



Kirchhoff’s Voltage Law (KVL)“The sum of all voltages in a closed loop is equal zero.”

+

V1

-

+

V3

-

+ V2 -

V1 – V2 – V3 = 0 or -V1 + V2 + V3 = 0



Voltage & Current Sources Combination

3V

2V

1V

4V

6A 5A 4A 7A



SUGGESTIONS FOR USING KIRCHHOFF'S LAWS

1. Place a (+) sign next the long line of the battery symbol and a (-) sign next to the short line. Start choosing a direction for conventional current flow ( flow of positive charge )If you choose the wrong direction for the flow of current in a particular branch, your final answer for the current in that branch will be negative. The negative answer indicates that the current actually flows in the opposite direction.

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• 2. Assign a direction to the circuit in each independent branch of the circuit. Place a positive sign on the side of each resistor where the current enters and a negative sign on the side where the current exits, e.g.; This indicates that a drop in potential occurs as the current passes through the resistor .


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• Notice how the directions of the currents are labeled in each branch of the circuit


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• 3. Select a JUNCTION POINT and apply the junction rule, e.g., at point A in the diagram:

The junction rule may be applied at more than one junction point. In general, apply the junction rule to enough junctions so that each branch current appears in at least one equation.


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• 4. Apply Kirchhoff’s loop rule by first taking note whether there is a gain or loss of potential at each resistor and source of emf as you trace the closed loop. Remember that the sum of the gains and losses of potential must add to zero.


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For example, for the left loop of the sample circuit start at point B and travel clockwise around the loop. Because the direction chosen for the loop is also the direction assigned for the current, there is a gain in potential across the battery (- to +), but a loss of potential across each resistor (+ to -).


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Following the path of the current shown in the diagram and using the loop rule, the following equation can be written:


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The direction taken around the loop is ARBITRARY. Tracing a counterclockwise path around the circuit starting at B, as shown in the diagram, there is gain in potential across each resistor to (- to +) and a drop in potential across the battery (+ to -). The loop equation would then be:


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• Multiplying both sides of the above equation by - 1 and algebraically rearranging, it can be shown that the two equations are equivalent. Be sure to apply the loop rule to enough closed loops so that each branch current appears in at least one loop equation. Solve for each

branch current using standard algebraic methods.“Solve simultaneous equations”


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Loop Current AnalysisExampleFind the current flowing in each branch of this circuit.


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R1

R2

V

+VR1

-

+VR2

-

Voltage Divider

VRR

RV

VRR

RV

R

R

21

221

1

2

1

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Example: Voltage divider

10V

4K

2K2K

+V1-

Find V1

VVKK

KV 333.310

42

21

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Current Divider

R2R1I

IR1 IR2

IRR

RI

IRR

RI

R

R

21

121

2

2

1

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Example: Current Divider

4K2K1mA

1mA

0V

1.333V 1.333V

0V

1.333V

0V

0.333mA

0.667mA

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Dependent Source

The amount of voltage (current) supplied depends on other voltage (current).

• Dependent Voltage Source

• Dependent Current Source

+

-

4Ix

2Vx

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Example

+-

5V

4Ω 1V 2Ω

3Vx

+ Vx -

I

Find I

AI

I

III

IVxVxII

222.0

418

0122145

)4(,032145

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When a direct current of I amperes is flowing in anelectric circuit and the voltage across the circuit isV volts, then

Unit of Energy is joule, when dealing with large amounts of energy, the unit used is the kilowatt hour (kWh)

Power and Energy

Power, in watts P = VIElectrical energy = Power * Time

= VIt joules

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Ques1. A source e.m.f. of 5V supplies a current of 3A for 10 minutes. How much energy is provided in this time?

Ques2. An electric heater consumes 1.8 MJ when connected to a 250V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply.

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Network Theorems


This chapter introduces important fundamental theorems of network analysis. They are the

Superposition theoremThévenin’s theoremNorton’s theoremMaximum power transfer theorem


Superposition Theorem

Used to find the solution to networks with two or more sources that are not in series or parallel.

The current through, or voltage across, an element in a network is equal to the algebraic sum of the currents or voltages produced independently by each source.

Since the effect of each source will be determined independently, the number of networks to be analyzed will equal the number of sources.


Superposition Theorem

The total power delivered to a resistive element must be determined using the total current through or the total voltage across the element and cannot be determined by a simple sum of the power levels established by each source.


Steps for Superposition Theorem

Remove all the sources except the one under consideration

Currents in various resistors and their voltage drops due to this single source are calculated.

This process is repeated for other sources taken one at a time.

Finally algebraic sum of currents and voltage drops over a resistor due to different sources is taken.


Select one energy source. Remove all other sources by

replacing voltage sources with a short while retaining any internal source resistance.

Replacing current sources with an open while retaining any internal resistances.

Repeat steps 1 through 2 for each other source individually.

Algebraically add the contributions of each voltage or current.


Superposition TheoremExample





Thévenin’s Theorem

Any two-terminal dc network can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.


Thévenin’s Theorem

Thévenin’s theorem can be used to:Analyze networks with sources that are not in

series or parallel.Reduce the number of components required to

establish the same characteristics at the output terminals.

Investigate the effect of changing a particular component on the behavior of a network without having to analyze the entire network after each change.


Procedure to determine the proper values of RTh and ETh

1. Remove that portion of the network across which the Thévenin equation circuit is to be found. In the figure below, this requires that the load resistor RL be temporarily removed from the network.


2. Mark the terminals of the remaining two-terminal network. (The importance of this step will become obvious as we progress through some complex networks.)

RTh:3. Calculate RTh by first setting all sources to zero

(voltage sources are replaced by short circuits, and current sources by open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.)


ETh:4. Calculate ETh by first returning all sources to

their original position and finding the open-circuit voltage between the marked terminals.


Conclusion:5. Draw the Thévenin

equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. This step is indicated by the placement of the resistor RL between the terminals of the Thévenin equivalent circuit.


Thevenin's TheoremExample


remove the load. remove the voltage source, VS1 and replace with a short.

We can now calculate the current through the resistor R3.


VTH Calculation


We are now ready to solve for RTH.

Remove all voltages sources

Replace them with a short, and open all current sources.



Norton’s Theorem

Norton’s theorem states the following: Any two-terminal linear bilateral dc network

can be replaced by an equivalent circuit consisting of a current and a parallel resistor.

RLRNIN


The steps leading to the proper values of IN and RN.

Preliminary steps:1. Remove that portion of the network across

which the Norton equivalent circuit is found.

2. Mark the terminals of the remaining two-terminal network.


Norton’s Theorem

Finding RN:3. Calculate RN by first setting all sources to zero

(voltage sources are replaced with short circuits, and current sources with open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.) Since RN = RTh the procedure and value obtained using the approach described for Thévenin’s theorem will determine the proper value of RN.


Norton’s Theorem

Finding IN :4. Calculate IN by first returning all the sources to

their original position and then finding the short-circuit current between the marked terminals. It is the same current that would be measured by an ammeter placed between the marked terminals.

Conclusion:5. Draw the Norton equivalent circuit with the

portion of the circuit previously removed replaced between the terminals of the equivalent circuit.


Example

3Ω

2Ω

10V

10Ω

RL

2Ω

Find Norton’s equivalent circuitand find the current that passes through RL when RL = 1Ω


3Ω

2Ω

10V

10Ω

2Ω

Isc

Find In

Find R total

Find I total

Current divider

4.4123

1232)210(||32

AR

VI 27.2

4.4

10

AISC 45.027.2123

3


3Ω

2Ω

10V

10Ω

2Ω

3Ω

2Ω 10Ω

2Ω

Short voltage source

RTH

2.13

232

3210

23||210THR

Find Rn


Norton’s equivalent circuit

If RL = 1Ω, the current is A418.045.012.13

2.13

RL13.20.45


Maximum Power Transfer Theorem

The maximum power transfer theorem states the following: A load will receive maximum power from a

network when its total resistive value is exactly equal to the Thévenin resistance of the network applied to the load. That is,

RL = RTh


RL

RS

VS

+

-

24

2

22

2

)(

)(2)(0

)(

SLS

LSLLS

L

L

SLS

LLLL

LS

SL

vRR

RRRRR

dR

dP

vRR

RRiP

RR

vi


Maximum Power Transfer Theorem

For loads connected directly to a dc voltage supply, maximum power will be delivered to the load when the load resistance is equal to the internal resistance of the source; that is, when:

RL = Rint


1.What is RL for maximum power transfer?

2.What is maximum power transfer delivered to RL?


RL = 12 ohms, P = 102 watts


C B

AA

C B

S

Ra

RbRc

--

- - - --

-R1

R2 R3

Delta-star transformation


Delta-star transformation

321

21

321

13

321

32

RRR

RRRc

RRR

RRRb

RRR

RRRa


Star-delta transformation

Rc

RaRbRbRaR

Rb

RcRaRaRcR

Ra

RbRcRcRbR

3

2

1


Example

4Ω

2Ω

1Ω

5Ω

3Ω

3/2Ω

2Ω

3/8Ω

5Ω

1/2Ω

13/2Ω19/8Ω

1/2Ω

159/71Ω

ENGG ZC111-L3

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