Lab #10
Kirchhoff’s Voltage
ECET 201003
ABSTRACT
Kirchhoff's Voltage Law states that the algebraic sum of voltages around a closed path is equal to zero. With regard to the equation it is:
or
These equations are essential for understanding the concept of measuring the voltages, it should be possible to verify this relationship. In performing an experiment of this nature, it should be remembered that each measurement is subject to some error. Thus, inconsistencies are expected within the calculations vs the data received from the digital multimeter.
INTRODUCTION
The general principle for series circuits stating that individual voltage drops add up to the total applied voltage, but measuring voltage drops in this manner reveals another facet of this principle: that the voltages measured as such all add up to zero. It doesn’t matter which point we start at or which direction we proceed in tracing the loop; the voltage sum will still equal zero. Kirchhoff’s Voltage Law, denoted as KVL for short, will work for any circuit configuration at all simple or complex, series or parallel.
Objective
To Verify Kirchhoff’s Voltage Law for DC circuits.
MATERIALS
PROCEDURE
RESULTS
Table 101 (Vs = 10 V)
R1 
R2 
R3 
R4 
Rt 
Calculated I (mA) = 0.625 
Measured Resistance values (kohms) 
1.2 
3.3 
4.7 
6.8 
16.0 
Calculated Voltages 
V1 
V2 
V3 
V4 
Vsum 
0.75 
2.0625 
2.9375 
4.25 
10.0 

Measured Voltages 
V1 
V2 
V3 
V4 
Vsum 
0.763 
2.117 
2.998 
4.310 
10.20 
Table 102 (Vs=10V)
R1 
R2 
R3 
R4 
Rt 
Calculated I (mA) = 0.529 
Measured Resistance Values (kohms) 
2.0 
2.2 
4.7 
10.0 
18.9 
Calculated Voltages 
V1 
V2 
V3 
V4 
Vsum 
1.058 
1.1638 
2.4863 
5.29 
9.9981 

Measured Voltages 
V1 
V2 
V3 
V4 
Vsum 
0.57 
1.25 
2.69 
5.68 
10.19 
Calculations for Table 101: Calculated I:
I= Vs(Rt) = 10V/(1.2 + 3.3 +4.7 + 6.8 kohms) = 0.625 mA
Calculated Voltages: Vx=I x Rx
V1= 0.625 mA x R1 = 0.625 mA x 1.2 kohms = 0.75 V
V2 = 0.625 mA x 3.3 kohms = 2.0625 V
V3 = 0.625 mA x 4.7 kohms = 2.9375 V
V4 = 0.625 mA x 6.8 kohms = 4.25 V
Vsum = 0.75 + 2.0625 + 2.9375 + 4.25 = 10.2 V
Vsource = 10V, Vsum very close to Vsource
Calculations for Table 102:
I= Vs(Rt) = 10V/(2.0 + 2.2 +4.7 + 10.0 kohms) = 0.529 mA
Calculated Voltages: Vx=I x Rx
V1= 0.529 mA x R1 = 0.529 mA x 2.0 kohms = 1.058 V
V2 = 0.529 mA x 2.2 kohms = 1.1638 V
V3 = 0.529 mA x 4.7 kohms = 2.4863 V
V4 = 0.529 mA x 10.0 kohms = 5.29 V
Vsum = 0.75 + 2.0625 + 2.9375 + 4.25 = 9.9981 V
Vsource=10V, Vsum very close to Vsource again.
Differences due to rubber on the circuit’s wires.
Discussion Questions:
CONCLUSION
Overall, this lab helped the group understand that there is a certain amount of error when it comes to calculating the values of the voltages due to certain resistances. Furthermore, it also helped the group understand the difference between a circuit board in parallel and in series. It has also helped the group involve Ohm’s Law with regard to the equations and how they would be applied in a scenario where the error of calculated voltage and actual voltage is involved.
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