Download Electrical Networks

3.4 Electrical networks
One application of linear equations is to electrical networks containing voltage sources
and resistors. Here is a simple electrical network.
Example 1. This network consists of three lines that are connected at two nodes. The
positive direction in each line is indicated by an arrow and the currents in the lines are
labeled i1, i2 and i3. Each line has a
node #1
voltage source with the positive and
negative terminals indicated by +
i1
i2
i3
and -. The voltage of each voltage
+
sources are labeled E1, E2 and E3.
E1 = 5v
E2 = 10v
E3 = 8v
Each line has a resistor with
+
+
resistances labeled by R1, R2 and R3.
line #1
line #2
line #3
In this network values are given for
R1 = 4Ω
R2 = 2Ω
R3 = 6Ω
the voltages and resistances and the
problem is to find the currents.
node #2
By way of background, here is some
information on these three types of electrical components.
i. Lines: conducting paths through which current can flow. Current is the flow
of charged particles in a line, mostly electrons. The current i in the line
is the rate at which charge is crossing a cross section in the line.
Positive charges moving in the positive direction in the line count
positively and those moving in the opposite direction count negatively.
For negative charges it is just the reverse. To determine the sign of the
current, it is necessary to choose a positive direction in the line. Charge
is measured in coulombs and current is measured in amperes. One
ampere of current flowing in a wire means there is a net flow of one
coulomb of charge past any cross section in one second. The different
lines are connected at nodes.
ii. Voltage sources:
these include batteries, power supplies and generators. They
create electrical fields that cause charged particles in a line to
move and hence produce a current in the line. Voltage
sources have two terminals, one is the positive terminal
designated by + and the other the negative terminal designated
by -. The voltage E of a voltage source is a measure of the
strength of the voltage source. More precisely, it is the
change in electrical potential as one goes from the negative
terminal to the positive terminal. More generally, for every
pair of points p and q in space there is a potential difference as
one goes from p to q. Typically, as p or q moves along lines,
the potential difference between p and q doesn’t change
except when p or q move through a circuit element such as a
3.4 - 1
resistor. Voltage is measured in volts. If V is the potential
difference between two points, then the electric field transfers
an amount of energy equal to QV to a particle with charge Q
that moves between these two points.
iii. Resistors:
these impede the flow of current in a line. We shall only consider
linear resistors that obey Ohm’s law which says V = iR where V is
the drop in potential as one goes through the resistor in the direction
of the current, i is the current in the line and R is a constant called
the resistance of the resistor. Resistance is measured in Ohms. A
resistor with resistance of R ohms has the property that in order to
make a current of i amperes flow through the resistor it is necessary
to supply a potential difference of iR to the two ends of the resistor.
In order to get equations that allow us to find the currents in the lines, we use two
electrical laws, namely Kirchoff’s current law and Kirchoff’s voltage law.
Kirchoff’s current law states the following.
At any node, the sum of the currents going into the node is the sum of the currents
going out of the node.
or
in’s = out’s
Kirchoff’s current law is just a restatement of the principle of conservation of charge at
the node. If we apply this to node #1 of the above network we get the equation
i3 = i1 + i2
or
i1 + i2 - i3 = 0
We get the same equation if we apply the current law at node #2.
Kirchoff’s voltage law states the following.
In any loop of the network, the sum of the voltage rises in the loop is the sum of the
voltage drops in the loop.
or
up’s = down’s
3.4 - 2
Kirchoff’s voltage law is a consequence of the fact that the sum of the voltage changes as
you go around the loop is zero. To apply this, we have to pick out some loops in the
network and pick out directions to traverse the loops.
One loop is at the right. If we go around this loop
counterclockwise and apply the voltage law to this
loop we get the equation
i1
i2
-
E1 = 5v
+
E2 = 10v
+
line #1
E1 + i2R2 = i1R1 + E2
or
line #2
R1 = 4Ω
R2 = 2Ω
5 + 2i2 = 4i1 + 10
or
4i1 - 2i2 = -5
The second loop is at the right. If we go around this
loop counterclockwise and apply the voltage law to this
loop we get the equation
i2
i3
-
+
E2 = 10v
E2 + E3 = i2R2 + i2R2
+
line #2
or
10 + 8 = 2i2 + 6i3
R2 = 2Ω
or
E3 = 8v
line #3
R3 = 6Ω
2i2 + 6i3 = 18
There is another loop where we go around the entire circuit first going along line 1 and
then line 3. However the equation we get from applying the voltage law to this loop is
the sum of the other two loop equations, so it doesn’t yield any new information.
We now have the following three equations for i1, i2 and i3.
i1 + i2 - i3 = 0
4i1 - 2i2
= -5
2i2 + 6i3 = 18
1 1 -1 | 0
The augmented matrix is  4 -2 0 | -5. Subtracting 4 times row 1 from row 2 gives
 0 2 6 | 18 
1 1 -1 | 0
 0 -6 4 | -5. Interchanging rows 2 and 3 gives  01 21 -16 || 180 . Dividing row 2 by 2 gives
 0 2 6 | 18 
 0 -6 4 | -5
1 1 -1 | 0
 0 1 3 | 9 . Subtracting row 2 from row 1 and adding 6 times row 2 to row 3 gives
 0 -6 4 | -5
 10 01 -43 || 9-9 . Dividing row 3 by 22 gives  10 01 -43 || 9-9 . Adding 4 times row 3 to row 1
 0 0 22 | 49
 0 0 1 | 49/22
1 0 0 | -2/22
and subtracting 3 times row 3 from row 2 gives  0 1 0 | 51/22 . So the solution is
 0 0 1 | 49/22
i1 = -2/22, i2 = 51/22 and i3 = 49/22.
3.4 - 3
If one or more of the voltage source values is not given, then we have more unknowns
than equations.
node #1
Example 2. In Example 1 suppose
the value of E3 is not given, so one
has the network at the right.
i1
i2
-
+
E1 = 5v
Now the equation that one obtains
by applying the voltage law to the
loop at the right is
i3
E2 = 10v
+
E3
+
-
line #1
R1 = 4Ω
line #2
line #3
R2 = 2Ω
R3 = 6Ω
10 + E3 = 2i2 + 6i3
or
2i2 + 6i3 - E3 = 10
node #2
So one now has the following set of three equations for i1, i2, i3 and E3.
i1 + i2 - i3
= 0
4i1 - 2i2
= -5
2i2 + 6i3 - E3 = 10
1
Let's solve for i1, i2 and i3 in terms of E3.  4
0
R2  R3

 10
0
R3/22  R3

1 -1 0 | 0
2 6 -1 | 10 
-6 4 0 | -5
 10
0
R2/2  R2

0 -4 1/2 | -5
1 3 -1/2 | 5 
0 1 -3/22 | 25/22
 10
0
1 -1 0 | 0
-2 0 0 | -5
2 6 -1 | 10 
1 -1 0 | 0
1 3 -1/2 | 5 
-6 4 0 | -5
R1 + 4R3  R1
R2 - 3R3  R2

 10
0
R2 - 4R1  R2

R1 - R2  R1
R3 + 6R2  R3

 01
0
 10
0
1 -1 0 | 0
-6 4 0 | -5
2 6 -1 | 10 
0 -4 1/2 | -5
1 3 -1/2 | 5 
0 22 -3 | 25
0 0 -1/22 | -10/22
1 0 2/22 | 35/22 .
0 1 -3/22 | 25/22 
The corresponding equations are
i1
- (1/22)E3 = -10/22
i2
- (2/22)E3 = 35/22
i3 - (3/22)E3 = 25/22
So the formulas for i1, i2 and i3 in terms of E3 are
i1 = (1/22)E3 - 10/22
i2 = (2/22)E3 + 35/22
i3 = (3/22)E3 + 25/22
or
 i1 
 1/22 
 -10/22 
 i2  =  2/22 E3 +  35/22 
 i3 
 3/22 
 25/22 
These formulas would be useful if we set E3 to different values at different times. We
could plug the current value of E3 into these formulas and find i1, i2 and i3.
3.4 - 4