Transcript Lesson 4 Circuit Structure

Basic Laws of Electric Circuits
Nodes, Branches, Loops and
Current Division
Lesson 4
Basic Laws of Electric Circuits
Nodes, Branches, and Loops:



Before going further in circuit theory, we consider the
structure of electric circuits and the names given to various
member that make up the structure.
We define an electric circuit as a connection of electrical
devices that form one or more closed paths.
Electrical devices can include, but are not limited to,
resistors
capacitors
inductors
1
transistors
logic devices
switches
transformers
light bulbs
batteries
Basic Laws of Electric Circuits
Nodes, Branches, and Loops:
A branch: A branch is a single electrical element or device.





Figure 4.1: A circuit with 5 branches.
A node: A node can be defined as a connection point between
two or more branches.




2
Figure 4.2: A circuit with 3 nodes.
Basic Laws of Electric Circuits
Nodes, Branches, and Loops:



3
If we start at any point in a circuit (node), proceed through
connected electric devices back to the point (node) from
which we started, without crossing a node more than one time,
we form a closed-path.
A loop is a closed-path.
An independent loop is one that contains at least one element
not contained in another loop.
Basic Laws of Electric Circuits
Nodes, Branches, and Loops:

The relationship between nodes, branches and loops
can be expressed as follows:
# branches = # loops + # nodes - 1
or
B=L + N - 1


4
Eq. 4.1
In using the above equation, the number of loops are
restricted to be those that are independent.
In solving most of the circuits in this course, we will not
need to resort to Eq. 4.1. However, there are times when it
is helpful to use this equation to check our analysis.
Basic Laws of Electric Circuits
Nodes, Branches, and Loops:
Consider the circuit shown in Figure 4.3.
+
_
Figure 4.3: A multi-loop circuit
5
give the number of nodes
give the number of independent loops
give the number of branches
verify Eq. 4.1
Basic Laws of Electric Circuits
Single Node Pair Circuits: Current division.
A single node pair circuit is shown in Figure 4.4
I
+
V
I2
I1
R2
R1
_
Figure 4.4: A circuit with a single node pair.
We would like to determine how the current divides (splits)
in the circuit.
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Basic Laws of Electric Circuits
Single Node Pair Circuits: Current division.
I
I
I2
+
I1
V
R2
_
I  I1  I 2 
R1
+
V
Req
_
V
V

R1 R2
Eq. 4.2
I
V
Req
Eq. 4.3
Therefore;
R1  R2
1
1
1



Req
R1 R2
R1 R2
7
Eq. 4.4
Basic Laws of Electric Circuits
Single Node Pair Circuits: Current division.
From Eq. 4.4 we can write,
Req 
R1 R2
R1  R2
Eq. 4.5
Equation 4.5 is a very important expression. In words it
says that the equivalent of two resistors in parallel equals to
the product of the two resistors divided by the sum.
The equivalent resistance of two resistors in parallel is always
less than the smallest resistor.
8
Basic Laws of Electric Circuits
Single Node Pair Circuits: Current division.
In general, if we have N resistors in parallel as in Figure 4.5
Req
R1
R2
  
RN
Figure 4.5: Resistors in parallel.
1
1
1
1


. . .
Req
R1 R2
RN
9
Eq. 4.6
Basic Laws of Electric Circuits
Single Node Pair Circuits: Current division.
Back to current division: We can write from Figure 4.4;
IReq
IR2
V
I1 


R1
R1
R1  R2
In summary form;
I1 
10
IR2
R1  R2
I2 
IR1
R1  R2
The above tells us how a current I divides when fed into
two resistors in parallel. Important
Eq. 4.7
Basic Laws of Electric Circuits
Single Node Pair Circuits: Current division.
In general, if we have N resistors in parallel and we want to
find the current in, say, the jth resistor, as shown in Figure 4.6,
I
Ij
Req
R1
R2
  
Rj
  
RN
Figure 4.6: General case for current division.
Ij 
11
IReq
Rj
Eq. 4.8
Basic Laws of Electric Circuits
Current Division: Example 4.1
Given the circuit of Figure 4.7. Find the currents I1 and I2
using the current division.
10 A
I2
I1
4
12 
Fig 4.7: Circuit for Ex. 4.1.
By direct application of current division:
I1 
12
10(4)
 2.5 A
12  4
I2 
10(12)
 7.5 A
12  4
Basic Laws of Electric Circuits
Current Division: Example 4.2
Given the circuit of Figure 4.8. Find the currents I1 and I2
using the current division.
7
10 A
I2
I1
4
12 
Figure 4.8: Circuit for Ex. 4.2.
The 7  resistor does not change that the current
toward the 4 and 12 ohm resistors in parallel is 10 A.
Therefore the values of I1 and I2 are the same as in
Example 4.1.
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Basic Laws of Electric Circuits
Current Division: Example 4.3
Find the currents I1 and I2 in the circuit of Figure 4.9 using
current division. Also, find the voltage Vx
I
7
+
20 V
+
_
Vx
I2
4
I1
12 
Figure 4.9: Circuit for Ex. 4.3.
_
We first find the equivalent resistance seen by the 20 V source.
4(12)
Req  7 
 7  3  10 
12  4
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Basic Laws of Electric Circuits
Current Division: Example 4.3
We can now find current I by,
20 20
I 

 2A
Req 10
We now find I1 and I2 directly from the current division rule:
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I1 
2(4)
 0.5 A
12  4
I2 
2(12)
 1.5 A
12  4
Basic Laws of Electric Circuits
Current Division: Example 4.3
We can find Vx from I1x12 or I2x4. In either case we get Vx = 6 V.
I
7
+
20 V
+
_
Vx
I2
I1
4
12 
_
We can also find Vx from the voltage division rule:
Vx 
16
20(3)
 6V
73
Basic Laws of Electric Circuits
Current Division: Example 4.4
For the circuit of Figure 4.10, find the currents I1, I2, and I3
using the current division rule.
I3
10 
15 A
I1 
I2
(15)( Req )
4
,
I2 
I1
Figure 4.10: Circuit
4
for Example 4.4.
20 
(15)( Req )
20
,
I3 
(15)( Req )
10
,
1
1
1
1
1
1
1



 

 0.25  0.05  0.1  0.4 S
Req
R1 R2
R3 4 20
10
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Basic Laws of Electric Circuits
Current Division: Example 4.4
I1 
 15 2.5
4
  9.375 A
I2 
(15)(2.5)
  1.875 A
20
I3 
(15)(2.5)
  3.75 A
10
We notice that I1 + I2 + I3 = - 15 A
as expected.
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Basic Laws of Circuits
circuits
End of Lesson 4
Nodes, Branches, Loops, Current Division
Basic Laws of Electric Circuits
Current Division: Example 4.4
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Basic Laws of Electric Circuits
Current Division: Example 4.4
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