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Physics 102
Kirchhoff’s Laws
Maria Zavala
February 14, 2006
Abstract:
The purpose of this lab is to compare measured branch currents with predicted currents
by applying Kirchhoff’s circuit laws to multi-loop circuits.
Equipment:
● Resistors (100<R<1000Ω)
● Breadboard
● VOM
● Old Fashion Ammeter
● Batteries
● Wires
Procedure:
1. Choose three different values for R1, R2, and R3 between 100 and 1000Ω.
2. Construct circuit 1 with the three resistors and three power supplies as follows
(refer to Fig. 1 of appendix):
a. Connect the positive side of the first power supply (V1) to R1.
b. Connect R1 to a parallel series of R2 and R3.
c. Connect the positive side of V2 to R2.
d. Connect the positive side of V3 to the negative side of V2.
e. Connect R3 and V3 to V1.
3. Measure the voltage across each circuit element.
4. Measure the current of each circuit branch.
5. Construct circuit 2 with four resistors and three power supplies as follows (refer to
Fig. 2 of the appendix):
a. Connect the positive side of V1 to R1.
b. Construct a loop parallel series R2 and R3 and connect it to R1.
c. Connect R4 to R1.
d. Connect the positive side of V2 to R4.
e. Connect the positive side of V3 to the negative side of V2.
f. Connect the negative side of V3 to the negative side of V1.
6. Repeat steps 3 and 4 for circuit 2.
Data:
Circuit 1
Power
supply
V1
V2
V3
Voltage
(V)
6
3
9
Resistors
R1
R2
R3
Resistance
(Ω)
180
820
560
Voltage
(V)
4.03
5.01
12.21
Current
(mA)
Resistors
R1
R2
R3
R4
Resistance
(Ω)
180
120
560
820
Voltage
(V)
0.992
0.559
0.559
4.54
Current
(mA)
2
7
9
Circuit 2
Power
supply
V1
V2
V3
Voltage
(V)
6
3
9
Calculations:
For Circuit 1
The equations to find theoretical value of current (I) are as follow:
I1= I2+I3
+V1-I1R1-I2R2-V2-V3=0
+V3+V2+I2R2-I3R3=0
-I1+I2+I3=0
-I1(180Ω)-I2(820Ω)=-6V+3V+9V
I2(820Ω)-I3(560Ω)=-9V-3V
By solving for I the results for current in circuit 1 are as follow:
I1=2.20mA
I2=-7.80mA
I3=10.01mA
For Circuit 2
The equations to find the theoretical value of current (I) are as follow:
I1=I3
I2R2+I2R3=0
V1-I1R1-I3R4-V2-V3=0
6
0
0
6
I1-I3=0
I2(120Ω+560Ω)=0
-I1(180Ω)-I3(820Ω)=-6V+3V+9V
By solving for I the results for current in circuit 2 are as follow:
I1=-6.00mA
I2=0mA
I3=-6.00mA
% Error:
The formula to find percent error is as follows:
│ (Theoretical Value- Actual Value) / Theoretical Value│x 100% = %error
For circuit 1
The percent error for current is as follows:
│( 2.20-2.00)/2.20│x 100%= 9.10% error for I1,
10.3 % error for I2 and
10.1% for I3.
For circuit 2
The percent error for current is as follows:
I1=0%
I2=0%
I3=0%
Error Analysis:
The errors of this lab are mainly due to the quality of the equipment and the accuracy in
which the instruments are used. The percent error was very low about 10% or less
through out the experiment which indicates that the set up of the circuits was done
correctly. The percent error may be due to the quality of the resistors. It is possible that
they are too old and the Ohms are less than what they were originally. Another
possibility is when the current was measured with the ammeter the wires were not
contacting the ends of each branch thus giving the wrong measurement. The set up of the
equations to find the theoretical current can also give an error if the set up is done wrong.
The negative signs on the results for the theoretical values of the current are due to
choosing the wrong direction for the current. The negative sign is arbitrary and the
magnitude of the currents is correct.
Overall this experiment was simple but with great room for error.
Conclusion:
The values for the actual currents in circuit 1 are I1=2mA, I2=7mA and I3=9mA. And the
theoretical current I1=2.20mA, I2=7.80mA, and I3=10.01mA. The values for the actual
currents in circuit 2 are I1=6.0mA, I2=0mA, and I3=6.0mA and the theoretical values for
the current I1=6.0mA, I2=0mA, and I3=6.0mA.
Appendix:
Fig. 1: Circuit 1
Fig.2: Circuit 2
Grade = 100