Experiment 2
Network Laws
John Nosek
ENEE 206
Section 101
Lab Report 2
2/24/04
● Objective
To study electrical network laws including Kirchhoff’s Voltage Law, Kirchhoff’s Current
Law, Node Voltages, and Mesh Currents.
● Equipment
- DC Power Supply
- Digital Multimeter
- Breadboard
- 500 Ω Resistor
- 1 kΩ Resistor
- 2 x 2 kΩ Resistors
● Schematics
Fig. 1
R2
c
d
I
R1
R3
e
b
V6
V12
a
R1 = 500 Ω
R2 = 1 kΩ
R3 = 2 kΩ
V12 = 12 V DC
V6 = 6 V DC
Fig. 2
R4
I6
R1
R3
A
C
B
I5
R1 = 500 Ω
R2 = 1 kΩ
R3 = 2 kΩ
R4 = 2 kΩ
V12 = 12 V DC
V6 = 6 V DC
I4
I1
I3
R2
I2
V12
V6
D
Fig. 3
R1
R3
I
Ia
R1 = 500 Ω
R2 = 1 kΩ
R3 = 2 kΩ
V12 = 12 V DC
V6 = 6 V DC
Ib
R2
V12
V6
● Procedure
For Part A, the circuit in Fig. 1 was constructed and each resistor was measured
with a multimeter to determine its actual value. The multimeter was then used to
measure the voltages Vab, Vbc, Vcd, Vde, and Vea across the designated points on the
circuit schematic. Also, the current I around the circuit loop was measured. Kirchhoff’s
Voltage Law was then used to verify the voltages around the loop abcdea. Voltages Vca,
Vdb, and Vce were then measured with the multimeter and compared with the calculated
values obtained by taking two different paths around the circuit loop using the five
previous voltage measurements.
Part B involved constructing the circuit in Fig. 2 and measuring each branch
current. Kirchhoff’s Current Law was then verified at each node A, B, C, and D using
the branch currents I1, I2, I3, I4, I5, and I6.
Mesh currents were explored in Part C by constructing the circuit in Fig. 3. First,
the voltages across the voltage sources V6 and V12 were measured. Then the current
loops Ia and Ib were measured along with the branch current I. Using the Mesh Current
Method and the measured values of V6, V12, R1, R2, and R3, the values of Ia, Ib, and I
were calculated and compared with the measured values.
● Results
Part A
1)
R1 = 507.1 Ω
R2 = 982 Ω
R3 = 2036 Ω
2)
Vab = 12.00 V
Vbc = -2.591 V
Vcd = -5.015 V
Vde = -10.39 V
Vea = 6.00 V
I = -5.10 mA
3)
KVL
 V12  IR1  IR2  IR3  V6  0
 12  I (507.1)  I (982)  I (2036)  6  0
I (507.1  982  2036)  18
I  .00510
I = 5.1 mA
4)
Vca = -9.41 V
Vdb = 7.60 V
Vce = -15.40 V
5)
Vca = -9.41
Path 1
Vca  IR2  IR3  V6  0
Vca  (.0051)(982)  (.0051)( 2036)  6  0
Vca  9.3918  0
Vca  9.3918
Path 2
V6  IR3  IR2  Vca  0
6  (.0051)( 2036)  (.0051)(982)  Vca  0
 9.3918  Vca  0
Vca  9.3918
Both values are very to close.
Vdb = 7.60
Path 1
Vdb  IR3  V6  V12  0
Vdb  (.0051)( 2036)  6  12  0
Vdb  7.6164  0
Vdb  7.6164
Path 2
V12  V6  IR3  Vdb  0
12  6  (.0051)( 2036)  Vdb  0
7.6164  Vdb  0
Vdb  7.6164
Both values are extremely close.
Vce = -15.40
Path 1
V6  V12  IR1  Vce  0
6  12  (.0051)(507.1)  Vce  0
15.41379  Vce  0
Vce  15.41379
Path 2
IR1  V6  V12  Vce  0
(.0051)(507.1)  6  12  Vce  0
 15.41379  Vce  0
Vce  15.41379
Both values are extremely close
Part B
2)
I1 = 11.62 mA
I2 = -7.78 mA
I3 = -3.83 mA
I4 = -0.80 mA
I5 = -8.58 mA
I6 = 3.05 mA
3)
Node A
I1  I 5  I 6
11.62  8.58  3.05
3.04  3.05
KCL Verified
Node B
I2  I4  I5
 7.78  0.80  8.58
 8.58  8.58
KCL Verified
Node C
I3  I6  I4
 3.83  3.05  0.80
 0.78  0.80
KCL Verified
Node D
0  I1  I 2  I 3
0  11.62  7.78  3.83
0  0.01
KCL Verified
Part C
2)
V12 = 12.00 V
V6 = 6.00 V
3)
Ia = 8.60 mA
Ib = 0.81 mA
I = 7.79 mA
4)
Loop 1
V12  I a R1  ( I a  I b ) R2
12  I a (507.1)  I a (982)  I b (982)
12  I a (1489.1)  I b (982)
Loop 2
 V6  ( I b  I a ) R2  I b R3
 6  I b (982)  I a (982)  I b (2036)
 6  I b (3018)  I a (982)
Solving for Ia in terms of Ib
12  I b (982)  I a (1489.1)
I a  .008  I b (.66)
Solving for Ib
 6  I b (3018)  (.008  I b (.66))(982)
 6  I b (3018)  7.856  I b (648.12)
1.856  I b (3666.12)
I b  5.06  10  4
Ib = .506 mA
Solving for Ia
12  I a (1489.1)  (5.06  10 4 )(982)
12.49  I a (1489.1)
I a  .0084
Ia = 8.40 mA
Solving for I
Ia  Ib  I
8.40  .506  I
I  7.89
I = 7.89 mA
The calculated values come close to the measured values enough to prove the Mesh
Current Analysis Method but are not very exact. There is some slight mathematical error
and measurement error for the measure values.
● Conclusion
Each circuit setup in the experiment clearly showed Kirchhoff’s Laws in a
physical example. By taking measurements of values and computing the same values, it
was easy to see how Kirchhoff’s Laws relate to the circuits. Each example also used
extensively the Node Voltage and Mesh Current Analysis Methods to help see the laws in
action and how these analyzing methods are seen and utilized in a physical circuit.
Although for some examples the measured values and computed values were slightly
different, because of measurement and mathematical errors, it was easy to see the
relationship between the circuit model and the law being tested.
The experiments gave good experience is setting up different kinds of circuits and
how to properly take voltage and amperage measurements in them. Also, use of the
circuit analysis methods was reinforced in the lab. Finally, the relationship between
voltage, amperage, and resistance was explored even further by analyzing all three in an
electrical network.