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DET 101/3
Basic Electrical Circuit 1
DC CIRCUITS:
CHAPTER 2
RESISTIVE CIRCUITS




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
Series/Parallel Equivalent Circuits
Voltage Divider Rule (VDR)
Current Divider Rule (CDR)
Voltage and Current Measurements
Wheatstone Bridge
Delta (or Pi) and Wye (or Tee) Equivalent
Circuit
Series/Parallel Equivalent Circuits
 Most common connection found in circuit
analysis.
 Circuit simplifying technique.
 Several resistors are combined to represent a
single equivalent resistance.
 Equivalent resistance depends on two (2)
factors:
Type of connection
Point of terminals
Series Equivalent Circuit
 The equivalent resistance for any number of
resistors in series connection is the sum of each
individual resistor.
x
x
R1
R2
R3
R4
RN
y
Req
R5
y
N
Req  R1  R2  R3  R4  R5    RN   Rn
n 1
(2.1)
Series Equivalent Circuit
(Continued…)
Apparently the single equivalent resistor is
always larger than the largest resistor in
the series connection.
Series resistors carry the same current
thru them.
Voltage across each of the resistors
obtained using voltage divider rule
principle or Ohm’s law.
Parallel Equivalent Circuit
 The equivalent resistance for any number of resistors in
parallel connection is obtained by taking the reciprocal of
the sum of the reciprocal of each single resistor in the
circuit.
x
x
R1
R2
R3
R4
R5
RN
Req
y
y
1

1
1
1
1
Req   



   RN 
 R1 R2 R3 R4 R5

1

1
 
 n 1 Rn
N



1
(2.2)
Parallel Equivalent Circuit
(Continued…)
Apparently, the single equivalent resistor is
always smaller than the smallest resistor
in the parallel connection.
Voltage across each resistor must be the
same.
Currents thru each of them are divided
according to the current divider rule
principle.
Parallel Equivalent Circuit
(Continued…)
Special simplified formula if the number of
resistors connected in series is limited to
two elements i.e. N=2.
Req
R1 R2

R1  R2
(2.3)
**When just two resistors
connected in parallel the
equivalent resistance is simply
the product of resistances
divided by its sum.
Special Cases of Connections: Open
Circuit (O.C)
 An opening
exists
somewhere in
the circuit.
 The elements
are not
connected in a
closed path.
i=0A
R1
R2
Vs
R3
a
b
O.C: i = 0 A
KVL: Voc = Vs
Ohm’s Law: Rab = V/I = ∞
Special Cases of Connections: Short
Circuit (S.C)
Both of its terminal
are joint at one
Vs
single node.
 The element is
bypassed.
i
a
R1
i
0A
R2
R3
b
S.C: Rab = 0 
Ohm’s Law: i = Vs/(R1 + R3)
: Vsc = 0 V
Practice Problem 2.9
Q: By combining the resistors in Figure below,
find Req.
Practice Problem 2.10
Q: Find Rab for the circuit in Figure 2.39.
Practice Problem 2.11
Q: Calculate Geq in the circuit of Figure 2.41.
Voltage Divider Rule (VDR)
Whenever voltage has to be divided
among resistors in series use voltage
divider rule principle.
+ V1 R1
R2
R3
+ V3 -
+ V2-
Vs
R1
V1 
Vs
R1  R2
R2
V2 
Vs
R1  R2
VDR (Continued…)
In general, to find the voltage drop across
the nth resistor in the voltage divider circuit
configuration we use this formula:
Rn
Vn 
Vs
R1  R2  R3    RN
(2.4)
Where n = 1, 2, 3,.....N
Practice Problem 2.12
 Find V1 and V2 in the circuit shown in Figure 2.43. Also
calculate i1 and i2 and the power dissipated in the 12
and 40 resistors.
Current Divider Rule (CDR)
Whenever current has to be divided
among resistors in parallel, use current
divider rule principle.
is
i1
Vs
R1
R2
i1 
is
R1  R2
i2
R2
R1
i2 
is
R1  R2
CDR (Continued…)
Circuit with more than two branches…
is
Vs
i1
i2
i3
R1
R2
R2
G1
i1 
is
G1  G2
G2
i2 
is
G1  G2
• In general, for N-conductors the formula represents:
Gn
in 
is
G1  G2  G3    GN
(2.5)
n = 1, 2, 3…..N
Practice Problem 2.13
 Find (a)V1 and V2 (b) the power dissipated in
the 3 k and 20 k resistors and (c) power
supplied by the current source.
1 k
3 k
+
V1
-
10 mA
5 k
Figure 2.45
+
V2
-
20 k
Chapter 2, Problem 34
 Determine i1, i2, v1, and v2 in the ladder network in Fig.
2.98. Calculate the power dissipated in the 2-
resistor.
Chapter 2, Problem 36
 Calculate Vo and Io in the circuit of Fig. 2.100.
Voltage and Current
Measurements
To determine the actual and quantitative
behavior of the physical system.
Two most frequently used measuring
devices in the laboratories:
 Ammeter
Voltmeter
Ammeter
Must be placed in series connection with
the element whose current is to be
measured.
An ideal ammeter should have an
equivalent resistance of 0  and
considered as short circuit equivalent to
the circuit where it is being inserted.
Voltmeter
Must be placed in parallel connection with
the elements whose voltage is to be
measured.
An ideal voltmeter should have an
equivalent resistance of ∞  and
considered as open circuit equivalent to
the circuit where it is being inserted.
Meter Types
 Analog meters
Based on the d’Arsonval meter movements.
 Digital meters
More popular than analog meters.
More precision in measurement, less
resistance and can avoid severe reading
errors.
Measure the continuous voltage or current
at discrete instants of time called sampling
times.
Configuration of Voltmeter and
Ammeter In A Circuit
RA = 0 
R1
Vs
A
R2
V
RV = Inf
Wheatstone Bridge : Practical Application of
Resistance Measurement
Invented by a British professor, Charles
Wheatstone in 1847.
More accurate device to measure
resistance in the mid-range (1  to 1 M)
In commercial models of the Wheatstone
bridge, accuracies about ± 0.1% are
achievable
Wheatstone Bridge (Continued…)
The bridge circuit
consists of
R1
R2
G
a
R3
Ig
b
x
V
+
V3
Figure A
-
+
Four resistors
A dc voltage source
A detector known asVs
galvanometer
(microampere range)
Rx
Balanced Bridge
If R3 is adjusted until the current Ig in the
galvanometer is zero the bridge its
balance state.
No voltage drop across the detector which
means point a and b are at the same
potential.
Implies that V3 = Vx when Ig = 0 A.
Balanced Bridge (Continued…)
Applying the voltage divider rule (VDR):
R3
V3 
Vs
R1  R3
and
Rx
Vx 
Vs
R2  Rx
Balanced Bridge (Continued…)
Since no current flows through the
galvanometer,
V3  Vx

R3
Rx
Vs 
Vs
R1  R3
R2  Rx
R2 R3  Rx R1
hence
R2
Rx 
R3
R1
(2.6)
Example 1
 The galvanometer shows a zero current through
it when Rx measured as 5 k. What do you
expect to be the value of the adjustable resistor,
R3? Show your derivation in getting the formula.
2 k
Ig
G
V3
V
-
+
R3
b
x
a
+
Vs
2k5
Rx
Exercise 1
 The bridge in Figure A is energized by 6V dc
source and balanced when R1 = 200, R2 =
500 and R3 = 800.
 (a) What is the value of Rx?
 (b) How much current (in miliamperes) does the
dc source supply?
 (c) Which resistor absorbs the least power and
which absorbs the most? How much?
Unbalanced Bridge
 To find Ig when the Wheatstone bridge is
unbalanced, use Thevenin equivalent circuit
concept to the galvanometer terminals.
 Assuming Rm is the resistance of the
galvanometer yields,
Ig
Vth

Rth  Rm
(2.7)
Delta (or Pi) and Wye (or Tee)
Equivalent Circuit
Stuck with neither series nor parallel
connection of the resistors in part of a
circuit.
Simplify the resistive circuit to a single
equivalent resistor by means of threeterminal equivalent circuit.
Wye/Tee Circuit
Same type of connections
R1
a
b
R1
R2
a
R2
b
R3
R3
c
c
(a)
Wye
c
(b)
Tee
Delta/Pi Circuit
Same type of Connections
Rc
Rc
a
b
Rb
a
Ra
Rb
Ra
c
c
(a)
b
Delta
c
(b)
Pi
Delta-to-Wye and Wye-to-Delta
Transformation
Remember that before and after
transformation using either Wye-to-Delta
or Delta-to-Wye, the terminal behavior of
the two configurations must retain.
Then only we can say that they are
equivalent to each other.
Special Case of -Y Transformation
 A special case occur when R1 = R2 = R3 = RY or
Ra = Rb = Rc =R under which the both networks
are said to be balanced. Hence the
transformation formulas will become:
RY = R/3
or
R = 3RY
 By applying Delta/Wye transformations, we may
find that this final process leads to series/parallel
connections in some parts of the circuit.
Delta to Wye Transform
 To obtain the equivalent resistances in the
Wye-connected circuit, we compare the
equivalent resistance for each pair of
terminals for both circuit configurations.
Rc ( Ra  Rb )
  connected : Rab  Rc || ( Ra  Rb ) 
Ra  Rb  Rc
Y  connected : Rab  R1  R2
Delta to Wye
Transform(Continued…)
To retain the terminal behavior of both
configurations i.e. R = RY
So that,
Rc ( Ra  Rb )
Rab 
 R1  R2
Ra  Rb  Rc
(2.9)
(2.8)
Ra ( Rb  Rc )
Rbc 
 R2  R3
Ra  Rb  Rc
Rb ( Ra  Rc )
Rca 
 R1  R3
Ra  Rb  Rc
(2.10)
Delta to Wye
Transform(Continued…)
 To obtain the resistance values for Y-connected
elements, by straightforward algebraic manipulation and
comparisons of the previous three equations gives,
Rb Rc
R1 
Ra  Rb  Rc
(2.12)
Ra Rb
R3 
Ra  Rb  Rc
(2.11)
Rc Ra
R2 
Ra  Rb  Rc
(2.13)
Wye to Delta Transform
 By algebraic manipulation, obtain the sum of all possible
products of the three Y-connected elements; R1, R2 and
R3 in terms of -connected elements; Ra, Rb and
Rc.(From Eq. (2.11 – 2.13)
Ra Rb Rc ( Ra  Rb  Rc )
R1 R2  R2 R3  R3 R1 
2
( Ra  Rb  Rc )
Ra Rb Rc

( Ra  Rb  Rc )
(2.14)
Wye to Delta
Transform(Continued…)
 Then we divide Eq. (2.14) by each of Eq. (2.11) to (2.13)
to obtain each of the -connected elements as to be
found variable in your left-side and its equivalent in Yconnected elements.
(2.14) / (2.11) :


Ra Rb Rc


 ( Ra  Rb  Rc ) 
R1 R2  R2 R3  R3 R1



Rb Rc
R1


(
R

R

R
)
b
c 
 a
Wye to Delta
Transform(Continued…)
(2.14) / (2.11) :
R1 R2  R2 R3  R3 R1
Ra 
R1
(2.15)
Using the same manner,
(2.14) / (2.12) :
R1 R2  R2 R3  R3 R1
Rb 
R2
(2.16)
(2.14) / (2.13) :
R1 R2  R2 R3  R3 R1
Rc 
R3
(2.17)
Wye-Delta Transformations
Delta -> Star
Star -> Delta
Rb Rc
R1 
( Ra  Rb  Rc )
Ra 
R1 R2  R2 R3  R3 R1
R1
Rc Ra
R2 
( Ra  Rb  Rc )
Rb 
R1 R2  R2 R3  R3 R1
R2
Ra Rb
R3 
( Ra  Rb  Rc )
Rc 
R1 R2  R2 R3  R3 R1
R3
Superposition of Delta and Wye
Resistors
 “Each resistor in the Yconnected circuit is the product
of the two resistors in two
adjacent  branches divided
by the sum of the three 
a
resistors”
 “Each resistor in the connected circuit is the sum of
all possible products of Y
resistors taken two at a time
divided by the opposite Y
resistors”
Rc
b
Rb
R1
R2
Ra
R3
c
Practice Problem 2.15
Q: For the bridge circuit
in Fig. 2.54, find Rab
and i.
i
a
13 

10 

100 V

b
Figure 2.54
50 
Exercise 2
 Use -to-Y transformation to find the voltages
v1 and v2.
15 
1
24 V
+
V1
-
10 

+
V2
-

Exercise 3
 Find the equivalent resistance Rab in the
circuit below.










Exercise 4
 Find Rab in the circuit below.
9 k
9 k
9 k
9 k
9 k
9 k
9 k
a
b
9 k
9 k