EE 1105
Pre-lab 4 Nodal and Loop Analysis
INTRODUCTION
Several fundamental laws govern the operation of all electrical circuits. In this lab
three of these will be addressed: Kirchhoff's Voltage Law, Kirchhoff's Current Law,
and the Conservation of Power Law. These laws can be stated as follows:
1. Kirchhoff's Voltage Law: The sum of the voltages across the elements contained in
a loop of an electrical circuit equals zero.
2. Kirchhoff's Current Law: The sum of the electrical currents exiting the node is
equal to zero.
3. Conservation of Power Law: The power applied to a circuit must be fully used in
the circuit, i.e., the power applied to the circuit equals the power used in the circuit.
(a) Loop Analysis (Use of Kirchhoff's Voltage Law )
Consider the following circuit:
4Ω
5Ω
5V
5V
1Ω
4Ω
5V
5Ω
I1I2
I1
5V
1Ω
1
EE 1105
1) Loop analysis is used to obtain the voltages in a loop.
 Pick a loop and number it.
 Draw arrows in the loop to show the flow of the current I 1 voltage.
 Write the signs of the voltage source by assigning the positive number
to the resistor when the loop is entering the resistor.
 Write the equations using Kirchoff’s Voltage Law.
-5 + 4I 1 - 5(I 2 - I 1 ) = 0
-5 + 1I 2 + 5(I 2 - I 1 ) = 0
 The two loop equations for this circuit can be rewritten as
9I 1 - 5I 2 = 5
-5I 1 + 6I 2 = 5
 These equations can be solved to find that I 1 = 55/29 A or about 1.897
A and I 2 = 70/29 A or about 2.414 A. The voltages across the resistors
can now be found using Ohm's Law:
Voltage across 1Ω: 1∙I 2 = 70/29 ≈ 2.414 V
Voltage across 4Ω: 4∙I 1 = 220/29 ≈ 7.586 V
Voltage across 5Ω: 5(I 2 - I 1 ) = 75/29 ≈ 2.586 V
 To verify these voltage results, one could use Kirchhoff's Voltage Law
and add the voltages around the outer loop of the circuit (the voltages
across the two sources and the 1Ω and 4Ω resistors), i.e.,
-5 + 7.586 -5 + 2. 414 = 0.
(b) Nodal Analysis (Use of Kirchhoff's Current Law)
The above solutions could also be found using Kirchhoff's Current Law. However,
since we have already solved for the loop currents above, one can use Kirchhoff's
Current Law to verify the solution via a different approach. Recall that Kirchhoff's
Current Law states that the sum of the currents entering a point in the circuit must
equal the sum of the currents exiting the same point in the circuit.
B
4Ω
5V
5Ω
1Ω
5V
2
EE 1105
Thus, at the top center node (the point labeled B; this node is the point that touches
one side of the 4Ω resistor and one side of the 5Ω resistor)
 Find a node in the circuit in this case is represented by the dot B.
 The sum of the currents exiting the node is equal to zero.
I1
I 1 + I 2 + I 3 =0
I2 1 Ω
Node
4Ω
I3
5Ω
 Write the equations in the next format using the V=IR where at this
time you have to use the currents (I). I1= 5/4, I2=1/5 and I3=5/5.
Using the above answers, this means that
1.897 = 2.414 + (1.897 - 2.414)
(c) A Laboratory setup to verify Kirchhoff's Laws
The above solutions could also be determined/verified in the lab. One would need
to measure elemental currents and/or voltages. A circuit for this purpose could be
setup as follows:
4Ω
Ammeter
Ammeter
Voltmeter
5V
Ammeter
5V
Voltmeter
5Ω
1Ω
Voltmeter
3
EE 1105
Note that the meters are placed to match the polarity of each resistor. For current
meters (ammeters) this means that if the current enters the positive side of the
resistor it must also enter the positive side of the ammeter.
Remember, that the polarity labeling on elements (other than sources) indicates
guesses at the orientation required for a pair of meter probes to give a positive
reading on the meter. If it turns out that one guessed the polarity direction
incorrectly for a positive reading, the local current/voltage will simply end up being
read as a negative value instead of positive value. A negative reading thus means
that one guessed the wrong polarity assignment to get the positive reading.
Another way to record current and voltage values is to label the nodes (the points
that connect elements). For example, let the nodes be label as follows:
A
B
4Ω
Voltmeter
5V
5V
5Ω
Voltmeter
G
C
1Ω
Voltmeter
Thus, the voltage across the 4Ω resistor could be recorded as V AB , where the first
letter in the subscript indicates the position of the positive meter connection. If one
reverses the meter probes, then one would record the voltage V BA . If V AB is a
positive value then V BA will be negative and vice-versa.
(d) Confirmation of the Conservation of Power Law
Once one knows the power values of all the elements in a circuit, then one can
confirm that the power applied to the circuit equals the power used by the elements
in the circuit. The power applied to a circuit by one source can be determined using
P source = V source I source .
4
EE 1105
If there is more than one source, then the applied power can be determined by
adding the powers supplied by all the sources. Similarly, the power used by
elements (in this lab, the resistors) in a circuit can be determined using
PR = VRIR.
The total power used in the circuit can be determined by adding the powers used by
the individual elements in the circuit.
II. PRELAB
Compute the requested quantities for each circuit. Develop wiring diagrams in
sufficient detail to allow you to quickly hook up each circuit and make the desired
measurements.
Note the resistor values are not all standard resistor values. How can parallel and
series combinations be used here? Just two standard values are required to make each
of these nonstandard resistors.
330 Ω
1.
560 Ω
5V
𝑉𝑠𝑜𝑢𝑟𝑐𝑒 =
𝐼𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑃𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑉560Ω =
𝐼560Ω =
𝑃560Ω =
𝑉330Ω =
𝐼330Ω =
𝑉330Ω+560Ω =
𝐼330Ω+560Ω =
2.
5V
330 Ω
𝑃330Ω =
𝑃330Ω+560Ω =
560 Ω
5
EE 1105
𝑉𝑠𝑜𝑢𝑟𝑐𝑒 =
𝐼𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑃𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑉560Ω =
𝐼560Ω =
𝑃560Ω =
𝑉330Ω =
𝐼330Ω =
𝑉330Ω+560Ω =
3.
𝑃330Ω =
𝐼330Ω+560Ω =
𝑃330Ω+560Ω =
330 Ω
560 Ω
5V
280 Ω
𝑉𝑠𝑜𝑢𝑟𝑐𝑒 =
𝐼𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑃𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑉560Ω =
𝐼560Ω =
𝑃560Ω =
𝑉330Ω =
𝑉890Ω =
𝑉330Ω+560Ω =
𝑉330Ω+560Ω+890Ω =
4.
𝐼330Ω =
𝑃330Ω =
𝐼890Ω =
𝑃890Ω =
𝐼330Ω+560Ω =
𝐼330Ω+560Ω+890Ω =
𝑃330Ω+560Ω =
𝑃330Ω+560Ω+890Ω =
560 Ω
330 Ω
5V
280 Ω
6
EE 1105
𝑉𝑠𝑜𝑢𝑟𝑐𝑒 =
𝐼𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑃𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑉560Ω =
𝐼560Ω =
𝑃560Ω =
𝑉330Ω =
𝑉890Ω =
𝑉330Ω+560Ω =
𝑉330Ω+560Ω+890Ω =
𝐼330Ω =
𝑃330Ω =
𝐼890Ω =
𝑃890Ω =
𝐼330Ω+560Ω =
𝐼330Ω+560Ω+890Ω =
𝑃330Ω+560Ω =
𝑃330Ω+560Ω+890Ω =
5.
280 Ω
560 Ω
330 Ω
5V
𝑉𝑠𝑜𝑢𝑟𝑐𝑒 =
𝐼𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑃𝑠𝑜𝑢𝑟𝑐𝑒 =
𝑉560Ω =
𝐼560Ω =
𝑃560Ω =
𝑉330Ω =
𝑉890Ω =
𝑉330Ω+560Ω =
𝑉330Ω+560Ω+890Ω =
𝐼330Ω =
𝐼890Ω =
𝐼330Ω+560Ω =
𝐼330Ω+560Ω+890Ω =
𝑃330Ω =
𝑃890Ω =
𝑃330Ω+560Ω =
𝑃330Ω+560Ω+890Ω =
7