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DET 101/3
Basic Electrical Circuit 1
DC CIRCUITS:
CHAPTER 2
RESISTIVE CIRCUITS
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 R N 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
R N
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 G N
(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