Download Lecture 2 - UniMAP Portal

RESISTANCE
CIRCUITS
RESISTANCE CIRCUITS
 Series/Parallel
Resistor
 Circuit Voltage Dividers
 Circuit Current Dividers
 Measurement of Voltage and Current
 Wheatstone Bridge
 Equal Circuit Delta-Wye (Pi-Tee)
Series Resistor
 Equivalent
Resistance
Req = R1 + R2 + ……….+ RN
Parallel Resistor
 Equivalent
Resistance
1
1
1
1


 ............ 
Req R1 R2
RN
Req 
1
1
R1
1

R2
1
 ........ 
RN
For two parallel resistor circuits
Req  R1 R2
Req 
1
1
R1
1

R1R2

R1  R2
R2
CIRCUITS VOLTAGE DIVIDER
By using Ohm’s law
V2  R2 I
V
I 
R1  R2
Voltage on resistor R2
 V 
 R2 
  V 

V2  R2 
 R1  R2 
 R1  R2 
Circuit Current Dividers
From Ohm’s law
 R1R2 
 (1)
V  I1R1  I 2 R2  I 
 R1  R2 
From equation (1)
 R2 
 I
I1  
R

R
2 
 1
 R1 
 I
I 2  
R

R
2 
 1
Measurement of Voltage and
Current
 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.
Ammeter and Voltmeter configuration
WHEATSTONE BRIDGE
 Galvanometer,
G was connected
between parallel path to detect
equilibrium.
 We
found,
I3 R3 = Ix Rx
 Balance
(1)
condition, voltage at R1
and R2 also should be same,
I1 R1 = I2 R2
(2)
 When
no current go through
Galvanometer, assume that,
I1 = I 3
(3)
 Change I3
with I1 and change IX
with I2 .we have
I1 R3 = I2Rx
(4)
Rx 
R2 R3
R1
 Devide
have
 We
eq. (2) and eq. (4), we
R1 R2

R3 Rx
have, RX as
R2 R3
Rx 
R1
(5)
DELTA-WYE CIRCUITS(PI-TEE)
 Galvanometer
in Wheatstone bridge
circuit replace by resistor Rm ,
 Resistor
R1, R2 and Rm (or R3,
Rm and Rx) called delta
connection (∆) and also called pi
(π) .
Configuration of delta circuit
 Resistor
R1, Rm and R3 (or R2,
Rm and Rx) known as wye (Y)
connection. This configuration
also known as tee (T)
connection.
Configuration of wye circuit
∆ - Y equivalent circuit
 By
using series-parallel theory,
equivalent resistance in ∆ circuit
for every terminal pair
Rc ( Ra  Rb )
Rab 
 R1  R2
Ra  Rb  Rc
Ra ( Rb  Rc )
Rbc 
 R2  R3
Ra  Rb  Rc
Rb ( Rc  Ra )
Rca 
 R1  R3
Ra  Rb  Rc
 From
previous three equation, we
have
Rb Rc
R1 
Ra  Rb  Rc
Rc Ra
R2 
Ra  Rb  Rc
Ra Rb
R3 
Ra  Rb  Rc
Y - ∆ circuits. Y circuits change
to ∆ circuit and every path could be
found like below.
 For
R1R2  R2 R3  R3 R1
Ra 
R1
R1R2  R2 R3  R3 R1
Rb 
R2
R1R2  R2 R3  R3 R1
Rc 
R3
EXAMPLE 1

From previous circuit, calculate.
 v0 if RL not connected
 v0 if RL = 150kΩ
 Power absorbed by 25kΩ
resistor if terminal load was close
circuit.
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
3
p  VI  (200)(8 10 )
 1.6W
EXAMPLE 2
Calculate power that absorb by resistor
6Ω resistor
Answer
 Equivalent
resistor
Req  ( 4 6 )  1.6  4
 find
i0,
 16 
i0  10
  8A
 16  4 
 If
current that flows through 1.6Ω
resistor is i0, current that flows
through 4Ω and 6Ω resistor can
be calculated
 4
i6  8   3.2 A
 10 
 Power
that obsorb by 6Ω resistor
could be calculated as below,
p  I R  (3.2) (6)  61.44W
2
2
Example 3

Calculate current and power that
have been supplied by power supply
40V.
 We
can select to have above
delta connection (100, 125, 25Ω)
or below delta connection (40, 25,
37.5Ω) and change to Y
connection. Here we select above
delta connection.
 Insert
Y resistance in the circuit
100 125
R1 
 50
250
125  25
R2 
 12.5
250
100  25
R3 
 10
250
 Equivalent
resistance,
50  50
Req  55 
 80
100
 Circuit
Equivalent circuit
 Current
I and power absorb by
circuit
40
i
 0.5 A
80
p  40  0.5  20W
Example 3
Answer
 Equivalent
resistance
60 30  20
 Current
i30
at 30Ω resistor
(25)(75)

 15 A
125
 v0
v0  (15)( 20)  300V
 Total
drop voltage at resistor
v0  30i30  300  450
 750V

vg
v g  12(25)  750
v g  1050V
Example 4
Answer
 Equivalent
resistance
5 20  4
4  6  10
10 40  8
 Then
125
ig 
 12.5 A
82
i6 
(40)(12.5)

 10 A
50
 thus
(5)(10)
i0 
 2A
25
QUIZ 2
QUIZ 3