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RESISTIVE CIRCUIT
Lecture 2
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent
circuit
SERIES/PARALLEL RESISTOR
Resistors that are arranged in series:
R1
I
VS
R2
 V1 

 V2 
RN
 VN 

I
VS


Resistance equivalent
Req
Req = R1 + R2 + ……….+ RN
CURRENT IN SERIES
CIRCUIT
• Current in series circuit is same at all
circuit elements
I  I1  I 2  I N
VOLTAGE IN SERIES CIRCUIT
• Voltage (VT) in series circuit is the total of
voltage for each elements.
VT  V1  V2  ..  VN
Resistors that are arranged in
parallel:
IS

V
I1
I2
R1
R2
IN
RN

IS

V

Req
Resistance equivalent:
1
1
1
1


 ............ 
Req R1 R2
RN
Req 
1
1
R1
1

R2
1
 ........ 
RN
For circuits with two resistors in parallel:
Req  R1 R2
Req 
1
1
R1
 1
R1R2

R1  R2
R2
CURRENT IN PARALLEL
CIRCUIT
• Current in parallel circuits is the total current
for each of the circuit elements.
I  I1  I 2  ..  I N
VOLTAGE IN PARALLEL CIRCUIT
• Voltage (VT) for parallel circuits is the same
for all circuits elements
VT  V1  V2  VN
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent
circuit
VOLTAGE DIVIDER
I
V

_
R1
 V1 

V2

R2
Using Ohm law, we will get:
V2  R2 I
V
I 
R1  R2
Voltage at resistor R2:
 V 
 R2 
  V 

V2  R2 
 R1  R2 
 R1  R2 
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent
circuit
CURRENT DIVIDER
I
V

_
I1
I2
R1
R2
Using Ohm law,
 R1R2 

V  I1R1  I 2 R2  I 
 R1  R2 
(1)
From equation (1):
 R2 
 I
I1  
 R1  R2 
 R1 
 I
I 2  
 R1  R2 
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent
circuit
VOLTAGE AND CURRENT
MEASUREMENT
• An ammeter is an instrument designed to
measure current.
• It is placed in series with the circuit element
whose current is being measured.
• An ideal ammeter has an equivalent
resistance of 0Ω and functions as a short
circuit in series with the element whose
current is being measured.
• A voltmeter is an instrument designed to
measure voltage.
• It is placed in parallel with the element
whose voltage is being measured.
• An ideal voltmeter has an infinite
equivalent resistance and thus functions
as an open circuit in parallel with the
element whose voltage is being measured.
The configurations for an ammeter and
voltmeter to measure current and voltage
R1
A
Vs


R2
V
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent
circuit
WHEATSTONE BRIDGE
• The Wheatstone bridge circuit is used to
precisely measure resistance of medium
values, that is in the range of 1Ω to 1MΩ.
• The bridge circuit consists of four resistors, a
dc voltage source and a detector.
The Wheatstone bridge circuit:
R1
R2
Vs
R3
RX
• When the bridge is balanced:
i1  i 3
i 3R 3  i x R x
i2  ix
i1R1  i 2 R 2
• Combining these equation, gives
i1R 3  i 2 R x
• Solving these equation, yields
R3 Rx

R1 R 2
R2
Rx 
R3
R1
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent
circuit
DELTA-WYE (PI-TEE) CIRCUIT
• If the galvanometer in Wheatstone
bridge is replace with its equivalent
resistance Rm,
• The resistor R1, R2 and Rm (or R3, Rm dan Rx)
are referred as a delta (∆) interconnection. It
also is referred as a pi (π) interconnection
because the ∆ can be shaped into a π
without disturbing the electrical equivalance
of the two confiurations.
Delta configuration
• The resistors R1, Rm and R3 (or R2, Rm
and Rx) are referred as a wye (Y)
interconnection because it can be
shaped to look like the letter Y. The Y
configuration also referred as a tee (T)
interconnection.
Wye configuration
The ∆ - Y transformation
Rc ( Ra  Rb )
• Using series
R


R

R
ab
1
2
and parallel
Ra  Rb  Rc
simplifications
Ra ( Rb  Rc )
in ΔRbc 
 R2  R3
connected,
Ra  Rb  Rc
yield
Rb ( Rc  Ra )
Rca 
 R1  R3
Ra  Rb  Rc
• Using straightforward
algebraic
manipulation gives,
Rb Rc
R1 
Ra  Rb  Rc
Rc Ra
R2 
Ra  Rb  Rc
Ra Rb
R3 
Ra  Rb  Rc
• The expression
for the three Δconnected
resistors as
functions of three
Y-connected
resistors are
R1R2  R2 R3  R3 R1
Ra 
R1
R1R2  R2 R3  R3 R1
Rb 
R2
R1R2  R2 R3  R3 R1
Rc 
R3
EXAMPLE 1
•
Find the current and power supplied by
the 40V sources in the circuit shown
below.
• We can find this equivalent resistance
easily after replacing either the upper Δ
(100, 125, 25Ω) or the lower Δ (40, 25,
37.5Ω) with its equivalent Y.
• We choose to replace the upper Δ. Thus,
100  125
R1 
 50
250
125  25
R2 
 12.5
250
100  25
R1 
 10
250
• Substituting the Y-resistor into the
circuit,
• The equivalent circuit,
• Calculate the equivalent resistance,
50  50
Req  55 
 80
100
Simplification of the circuit,
• Then, the current and power values
are,
40
i
 0.5 A
50
p  40  0.5  20W
Example 2
a) Find no load value of vo.
b) Find vo when RL = 150kΩ
c) How much power is dissipated
in the 25kΩ resistor if the load
terminals are accidentally shortcircuited?
Answer:
a)
b)
 75k 
v0  200
  150V
 100k 
75k  150k
Req 
 50k
75k  150k
 50k 
v0  200
  133.33V
 75k 
c)
V 200
3
I 
 8 10 A
R 25k
2
3
p  I R  (200)(8 10 )
 1.6W
EXAMPLE 3
Find the power dissipated in the
6Ω resistor.
Asnwer
• Equivalent resistance
Req  ( 4 6 )  1.6  4
• current io,
 16 
i0  10
  8A
 16  4 
• Note that io is the current in the 1.6Ω
resistor.
• Use current divider to get current in
the 6Ω resistor,
 4
i6  8   3.2 A
 10 
• Then the power dissipated by the
resistor is
p  I R  (3.2) (6)  61.44W
2
2
EXAMPLE 4
Find the voltage of vo and vg.
Asnwer
• Equivalent resistance
60 30  20
• Current in resistor 30Ω
i30
(25)(75)

 15 A
125
• Voltage v0
v0  (15)( 20)  300V
• Total voltage at the resistor
v0  30i30  300  450
 750V
• Voltage vg
v g  12(25)  750
v g  1050V
EXAMPLE 5
Find the current of ig and.io
Answer
• Equivalent resistance:
5 20  4
4  6  10
10 40  8
• Current values,
125
ig 
 12.5 A
82
i6 
(40)(12.5)

 10 A
50
• Thus,
(5)(10)
i0 
 2A
25
EXAMPLE 6
• Determine the value of io