First Year Experiment 5
Modules CY1B2, SE1B2 D.C. Circuit Theorems
Circuit theory is an important part of electronics, and this experiment investigates some simple circuit theory. In
particular, Ohm's Law, Kirchhoff's Laws, Thevenin's theorem and the Principle of Superposition are all used to
analyse a bridge circuit and measurements taken to test the validity of these theories.
Introduction
The term DC original meant direct current or directly coupled. It is now conventionally treated as a single word
implying voltages and currents that are not changing with time to a significant extent. Thus it is permissible to
speak of DC voltage or DC current. There is no such thing, however, as DC circuit theory. Ohm's Law,
Kirchhoff's Laws, Thevenin's theorem and the Principle of Superposition apply equally to DC signals as to
signals that change, so called AC signals. Read through the theory below, asking a demonstrator if you have
questions, and then do the experimental work as specified on the last page.
Basic Theory
Voltage is potential, the amount of work needed to transport a unit charge from one place to another in a circuit.
This is independent of the path followed, so the voltage around any complete path must be zero.
Current is a flow of charge, which is conserved. Thus the total current flowing into any point, or node, in a circuit
must equal the total current flowing out of that node.
These two requirements are conventionally known as Kirchhoff's Laws.
Current Law: The algebraic sum of currents flowing into a circuit node is zero.
Voltage Law: The algebraic sum of the voltages around a closed path (or loop or mesh) of a circuit is zero.
In this experiment, a circuit involving a voltage source and resistors is used. Ohm's law can be used to determine
the relationship between the current flowing through a resistor and the voltage across it. Figure 1 shows a resistor
where the current through it is I, the voltage across it is V and the resistance of the resistor is R.
Figure 1 Resistor
If I flows THROUGH resistor, voltage developed ACROSS it: Ohm's law says: V = IR
If V applied ACROSS resistor, current I flows THROUGH it: Ohm's law says: I = V/R
Potential Divider
Consider the so called potential divider circuit in figure 2. We want to determine the value of Vo.
Figure 2. Potential Divider Circuit
If we assume Io = 0, then at the node between the two resistors, Kirchhoff's current law states:
I1 = I2 + Io = I2.
; because we have assumed that Io = 0
Kirchhoff’s voltage law around the loop including the power source, VI, states that:
; because I1 = I2
VI = I1 * R1 + I2 * R2 = I2 * (R1 + R2)
Ohm's Law applied to the resistor R2 states that
VO = I2 * R2 or I2 = Vo/R2
V
R + R2
Therefore
VI = O * (R1 + R 2 ) = VO * 1
R2
R2
So (remember this)
VO =
R2
VI
R1 + R 2
However, if Io is not zero, then the above is not true, and the equations become more complicated. We will
investigate a bridge circuit where the above can be applied, and extended to cope with Io not being zero.
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Thevenin's Theorem
Thevenin's Theorem states that, as far as any external connection is concerned, a network of voltage (and current)
sources and resistors can be replaced by a voltage source equal to the open circuit voltage and a resistance equal
to the ratio of the open circuit voltage to a short circuit current.
Thus the Potential Divider circuit can be replaced by a circuit of the form shown in figure 3. We need VT & RT.
Figure 3 Thevenin Equivalent
For the potential divider, the Thevenin voltage, VT, is that we have found before, namely VT =
R2
VI
R1 + R 2
The Thevenin resistance, RT, is found by replacing the voltage source by a short circuit and determining the
resistance as seen by looking into the outputs. Thus the potential divider looks like that shown in figure 4.
Figure 4 Finding RT
Looking into the circuit there are two resistors in parallel, thus the Thevenin resistance is:
R R
RT = 1 2
R1 + R 2
Bridge Circuit
In this experiment, the so called Bridge circuit shown in figure 5, is to be investigated. Initially, the circuit will be
as shown, though later an extra resistor will be added between points labelled A and B.
Figure 5 Bridge Circuit
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Note 4.7kΩ = 4.7 * 10 Ω = 4700Ω.
As the circuit stands, it can be considered to be two potential divider circuits, both with 30V applied, one
involving the resistors with values 10kΩ and 2.7kΩ, the other with resistors with values 4.7kΩ and 3.3kΩ. Thus,
the voltage at node A, VTA, can be found by:
VTA =
2.7 * 10 3
30 =
81 * 10 3
= 6.38V
10 * 10 3 + 2.7 * 10 3
12.7 *10 3
The voltage at node B, VTB, can be found in a similar manner.
VTA and VTB are the Thevenin voltages of the two potential divider circuits. It is also possible to find the
Thevenin resistances for these two circuits using the parallel resistances formula. Let these be called RTA and RTB.
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Thevenin Equivalent of bridge circuit with resistor
Figure 6 shows the bridge circuit with an extra resistor connected between points A and B.
Figure 6 Bridge Circuit with extra Resistor
The addition of this resistor means that the potential dividers are now ones where current flows from the midpoint
in the two resistors, that is IO is now not zero. The analysis of the circuit becomes more difficult. However, the
analysis can be made more easy using Thevenin Equivalent Circuits and Superposition.
Now, we have said that the two potential dividers can each be replaced by their Thevenin equivalent circuit. Thus
the above can be represented by the circuit shown in figure 7.
Figure 7 Thevenin Equivalent Bridge Circuit
Principle of Superposition - to analyse bridge circuit with resistor
The principle of superposition, which applies to linear systems of which the bridge circuit is a simple example, is
a powerful method of analysing systems. For the circuit in Figure 7 we have two voltage sources, VTA and VTB. If
we want to find the voltage at point A, VA, the superposition principle is used as follows. Work out the voltage at
A assuming that VTB is 0: we will call this voltage VA1. Work out the voltage at A assuming VTA is 0: call this
VA2. Then the actual voltage we want, VA, is simply VA = VA1 + VA2. The same applies to the voltage at B.
VA1, for instance, is simply found using the potential divider rule, where R1 = RTA and R2 is 10k + RTB:
10k + R TB
VA1 =
VTA
R TA + 10k + R TB
The voltages VA2, VB1 and VB2 are found in a similar manner, and hence VA and VB are found easily.
Alternative Method - Using Kirchhoff's laws
The alternative is to use Kirchhoff's Laws, which give six equations for the six currents (as shown in figure 8).
Using Kirchhoff's current laws:
at node C:
I1 = I2 + I3
at node A:
I3 + I4 = I6
at node B:
I2 = I4 + I5
Considering voltages around loops:
loop CADC: 30 = 10k I3 + 2.7k I6
loop CABC: 10k I3 = 4.7k I2 + 10k I4
loop DABD: 2.7k I6 + 10k I4 = 3.3k I5
After much work these equations can be solved,
the results being:
I1 = 6.20mA I2 = 3.92mA I3 = 2.27mA
I4 = 0.43mA I5 = 3.50mA I6 = 2.70mA
Figure 8 Bridge Circuit with currents given.
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Experimental Procedure
Write in your log book, the date you are doing the experiment and the objectives of the experiment.
The equipment you are provided with is a 30V stabilised power supply, a 10kΩ resistor, a Bridge Circuit and a
multimeter. Note these in your log book.
The following details the experimental tasks. Obey each task, performing all relevant calculations and
measurements: record all you do, the results you find, and comments on the results, in your log book.
Analysing Circuit without 10kΩ resistor.
Connect the 30V power supply to the bridge circuit, but leave the terminals A and B disconnected.
In the theory section a value for the voltage at node A is given. Write down in your log book what that voltage is,
and work out what the voltage should be at node B: put all working in your log book.
Use the multimeter to measure these voltages. Do they agree with what you expect?
These two voltages are the Thevenin equivalent voltages of the two potential divider circuits, VTA and VTB.
Work out and record the Thevenin resistances of these two potential divider circuits, RTA and RTB.
Disconnect the 30V supply.
Connect a wire (that is insert a short circuit) between points C and D of the bridge.
Use the multimeter to measure the resistance between points A and D, and that between points B and D.
Compare these with the Thevenin resistances RTA and RTB.
Comment on your results.
Analysing Circuit with the 10kΩ resistor.
Use the Principle of Superposition to calculate voltages VA1, VA2, VB1 and VB2 and hence work out VA and VB.
Connect the 10kΩ resistor to the bridge circuit, remove the short circuit and connect the 30V supply.
Use the meter to measure the voltages at points A and B. Compare these with your values of VA and VB.
Comment on the results.
Analysing Circuit using Kirchhoff's laws
Verify that the six values of currents given above are solutions to the six equations, allowing for rounding errors.
With the 10kΩ resistor still in place, set the multimeter to read current and measure the currents I1, I2, I3, I4, I5 and
I6. Compare these with those determined by the theory.
If you have time, use simultaneous equation techniques to solve the six equations to find the six currents.
Conclusion
Write in your log book what you have concluded from doing this experiment.
Appendix - MATLAB Session
For information, the following is the MATLAB session which can be used to solve the six equations. This is done
by forming two matrices A and b such that the six equations are modelled by one matrix equation A I = b, where
I is a matrix with the currents, and then using the command I = A\b to find the matrix I.
» A=[1 -1 -1 0 0 0; 0 0 1 1 0 -1; 0 1 0 -1 -1 0; 0 0 10 0 0 2.7; 0 -4.7 10 -10 0 0;
0 0 0 10 -3.3 2.7]
A =
1.0000
-1.0000
-1.0000
0
0
0
0
0
1.0000
1.0000
0
-1.0000
0
1.0000
0
-1.0000
-1.0000
0
0
0
10.0000
0
0
2.7000
0
-4.7000
10.0000 -10.0000
0
0
0
0
0
10.0000
-3.3000
2.7000
» b=[0;0;0;30;0;0]
b = 0
0
0
30
0
0
»I=A\b
I = 6.1974
3.9259
2.2716
0.4264
3.4995
2.6979
RJM 16/02/04
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