Electric Circuits I - Philadelphia University · Electric Circuits I Simple Resistive Circuit 1 Dr....

Electric Circuits ISimple Resistive Circuit

1

Dr. Firas Obeidat

Dr. Firas Obeidat – Philadelphia University

The equivalent resistance of any number of resistorsconnected in series is the sum of the individual resistances.

It is often possible to replace relatively complicated resistorcombinations with a single equivalent resistor withoutchanging all the current, voltage, and power relationships inthe remainder of the circuit.

2

series combination of N resistorsvs=v1+v2+v3+v4+…+vN

vs=R1i+R2i+R3i+R4i+…+RNi

vs=(R1+R2+R3+R4+…+RN)i

vs=Reqi

Req: equivalent resistor

Req=R1+R2+R3+R4+…+RN

Resistors in Series

Resistors in Parallel


The equivalent resistance of two parallel resistors is equal tothe product of their resistances divided by their sum.

It is often possible to replace relatively complicated resistorcombinations with a single equivalent resistor withoutchanging all the current, voltage, and power relationships inthe remainder of the circuit.

3

Parallel combination of N resistors

is=i1+i2+i3+i4+…+iN

is=�

Req

�� =�

��+�

��+�

��+�

��+ ⋯ +

�

��

�

��=�

��+�

��+�

��+�

��+ ⋯ +

�

��

Resistors in Parallel

Dr. Firas Obeidat – Philadelphia University 4

A parallel combination is routinely indicated by the followingshorthand notation

�� = ��‖��‖��‖⋯ ‖��

The special case of only two parallel resistors isencountered fairly often, and is given by

�� = ��‖��

�� =�� × �� +��

�� ≠�� × �� × �� +�� +��

Examples


Example: find Rab?

This 1.2 Ω resistor is in series with the 10 Ω resistor.

the 2Ω and 3Ω resistors are in parallel

Example: find Rab?

Answer: 19Ω

Examples


Example: find Req?

Examples


Example: find Req?

Answer: 10Ω

Example: find Rab?The 3Ω and 6Ω resistors are in parallelbecause they are connected to the same twonodes c and b.

The 12Ω and 4Ω resistors are in parallel since they are connected to the same two nodes d and b.

Also the 1Ω and 5Ω resistors are in series

Examples


Example: find Rbc ,Rbc ,and Rbc?

�� = � + � ‖�‖� ‖ � + � + � + � = �.��

�� = (� + � ‖�‖� + � + �)‖[� + �] + �� = ��

�� = ��

Example: find Req and i?

�� = �.� + � = �.��

� =��

�.�= �.�A

Examples


Example: determine the current i in the figure and the powerdelivered by the 80 V source.

We first interchange the element positions in the circuit.

combine the three voltage sources into anequivalent 90 V source, and the fourresistors into an equivalent 30 Ω resistance

��= −�� − �� + �� = −��

��= �� + � + � + � = ��

-90+30i=0⇒� =��

��=3A

The desired power in the voltage source 80V is

80V × 3A =240W

Examples


Example: Calculate the power and voltage of the dependentsource in the figure.

The two 6Ω resistors are in parallel and can be replacedwith a single 3Ω resistor in series with the 15Ω resistor.Thus, the two 6Ω resistors and the 15Ω resistor arereplaced by an 18Ω resistor.

Also 9Ω||18Ω=6 Ω

The controlling variable i3 depends on the 3Ω resistorand so that resistor must remain untouched.

Applying KCL at the top node .

−�� − � + �� +�

�= �(�)

Employing Om’s law.

� = �� (2)

Solve equation (1) and (2), then i3=(10/3)A

Thus, the voltage across the dependent source, v=3i3=10V

pD=v×0.9i3=10×0.9×10/3=30W

Voltage Division


Voltage division is used to express the voltage across one ofseveral series resistors.

vs=v1+v2=R1i+R2i =(R1+R2)i

� =�

�� +��

�� = �� =�

�� +��

�� =��

�� +��

�� =��

�� +��

So

Thus

Or

And the voltage across R1 is similarly

�� =��

�� +�� +�� + ⋯ +��

General result for voltage divisionacross a string of N series resistors

Current Division


Current division is used to express the current across one ofseveral parallel resistors.

�� =�

��=� ��‖��

=�

��

�� +��

Or

And similarly

For a parallel combination of Nresistors, the current throughresistor Rk is

The current flowing through R2 is

�� = ��

�� +��

�� = ��

�� +�� = �

��

��+��+��+ ⋯ +

��

Examples


Example: Determine vx in the circuitof shown figure.Combine the 6Ω and 3Ω resistors, replacing them with:

6×3/(6+3)=2Ω

Since vx appears across the parallel combination,the simplification has not lost this quantity.

�� = ��

� + �= ��

Example: Determine vx in the circuitof shown figure.

Answer: 2V.13

Examples


Example: write an expression for thecurrent through the 3 resistor in thecircuit of shown figure.

The total current flowing into the combination is

The desired current is given by current division:

��(�)= ��

� + �=�

��

Example: In the circuit of Fig.3.39, use resistance combinationmethods and current division tofind i1, i2, and v3.

Answer: 100mA, 50mA, 0.8V.

�(�)=��

� + �||�=��

� + �= ��

14

Examples


Example: For the circuit shown in thefigure, determine: (a) the voltage vo (b)the power supplied by the currentsource, (c) the power absorbed by eachresistor.

(a) The 6kΩ and 12k Ω resistors are in series so that their combined value is 6k+12k=18k Ω.

�� =��

�� + ��(��)= ��

Applying the current division techniqueto find i1 and i2

�� =��

�� + ��(��)= ��

The voltage across the 9-k and 18-k resistors is the same,and vo=9000i1=18000i2=180V.

15

Examples


Example: For the circuit shown in thefigure, determine: (a) the voltage vo (b)the power supplied by the currentsource, (c) the power absorbed by eachresistor.

(b) Power supplied by the source is.

�� = �� = �� × �� = �.��

(c) Power absorbed by the 12kΩ resistor is.

p= �� = �� = �� × �� = �.��

Power absorbed by the 6kΩ resistor is.

� = �� = �� = �� × �� = �.��

Power absorbed by the 9kΩ resistor is.

� =��

�=(��)�

��= �.�� = �� = �� × �� = �.��or

16

Examples


Example: Design the voltage divider of the figure such thatVR1=4VR2.

The total resistance is defined by.

�� =�

�=

��

�.��= ��

Since VR1=4VR2

R1=4R2

�� = �� + �� = �� + �� = ��

��=5kΩ

��=1kΩ

R1=4R2=4kΩ

Examples


Example: calculate the indicated currents and voltage offigure shown below.

Redrawing the network after combiningseries elements yields

�� =�

��,�,�||��=

��

��=��

��= ��

�� =(��||��,�)�

(��||��,��=

�.��×��

�.��=��

��.�= ��.��

�� =��

��||��,�=��.��

�.��= �.��

�� = �� + �� = �� + �.�� = �.��

Examples


Example: Determine the voltages V1, V2, and V3 for thenetwork of the figure shown below.

�� − �� − �� =0

For path (1)

�� = �� − �� =20V-8V=12V

�� − �� − �� =0

For path (2)

�� = �� − �� = �V-12V=-7V

�� + �� − �� =0

For path (3)

�� = �� − �� = �V-(-7V)=15V

Indicating that V2 has a magnitude of 7 V but a polarity opposite to that appearing in the figure.

Delta-Wye Conversion


Parallel and series combinations of resistors can often lead to asignificant reduction in the complexity of a circuit. There is anotheruseful technique, called delta-wye (Δ-Υ ) conversion that can be usedto simplify the circuit.

Two forms of the same network: (a) Y, (b) T.

Two forms of the same network: (a) Δ,(b) Π.

These networks (Δ,Υ) occur bythemselves or as part of a largernetwork. They are used in three-phase networks, electrical filters,and matching networks.

Delta to Wye Conversion


For terminals 1 and 2.

��(Y)=�� +��

��(Δ)=��||(�� + ��)

��(Y)= ��(Δ)Setting

��=�� + �� =��(��)

��(1)

Similarly

��=�� + �� =��(��)

��(2)

��=�� + �� =��(��)

��(3)

To obtain the equivalent resistances in the wyenetwork, we compare the two networks andmake sure that the resistance between eachpair of nodes in the Δ (or Π) network is thesame as the resistance between the same pairof nodes in the Y (or T) network.

Delta to Wye Conversion


Subtracting eq. (4) from eq. (2)

�� =��

�� + �� + ��(6)


�� =��

�� + �� + ��(7)


�� − �� =��(�� − ��)

�� + �� + ��(4)

Adding eq. (2) from eq. (4)

�� =��

�� + �� + ��(5)

Each resistor in the Ynetwork is the product ofthe resistors in the twoadjacent Δ branches,divided by the sum of thethree Δ resistors.

Wye to Delta Conversion


To obtain the conversion formulas for transforming a wye network to anequivalent delta network. Multiply eq.(5)&eq.(6), eq.(6)&eq.(7),eq.(7)&eq.(1). Then add all the three equation together.

�� +�� +�� =��(��+�� + ��)

(�� + �� + ��)�

=��

�� + �� + ��(8)

Dividing eq. (8) by each of equations (5), (6)and (7) leads to the following equations

�� =��

��(9)

�� =��

��(10)

�� =��

��(11)

Each resistor in the Δ network isthe sum of all possible products ofY resistors taken two at a time,divided by the opposite Y resistor.

Delta Wye Conversion


The Y and Δ networks are said to be balanced when

�� = �� = �� = �� , �� = �� = �� = ��

Under these conditions, conversion formulas become

�� =��

�or �� = 3��

Example: Obtain the equivalent resistance for the circuit in the figure below anduse it to find current i.

In this circuit, there are two Ynetworks and three Δ networks.Transforming just one of these willsimplify the circuit. If we convert theY network comprising the 5Ω, 10Ω,and 20Ω resistors, we may select




�� = 10Ω, �� = 20Ω, �� = 5Ω,

�� =��×��×��×��

��=��

��=35Ω

�� =��

��=��

��= 17.5Ω

�� =��

��=��

�= 70Ω

�� =��

��




Combining the three pairs of resistors in parallel, we obtain

70||30 =70× 30

70+ 30= 21Ω

12.5||17.5 =12.5 × 17.5

12.5 + 17.5= 7.292Ω

15||35 =15× 35

15+ 35= 10.5Ω

�� = (7.292 + 10.5)||21 =17.792× 21

17.792+ 21= 9.632Ω

� =��

=120

9.632= 12.458Ω

Electric Circuits I - Philadelphia University · Electric Circuits I Simple Resistive Circuit 1 Dr....

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