Download File - Martin Ray Arcibal

Martin Ray A. Arcibal
5° AP Fiziks
28 April 2013
Mr. Tillay
Kirchhoff’s Laws: Voltage and Current in Circuits
(Current Sensor, Voltage Sensor)
1. Purpose
The purpose of this lab is to validate Kirchhoff’s two laws, the junction rule and the loop
rule, when applied to circuits.
2. Hypothesis
According to Kirchhoff’s junction rule, the amperage of a circuit just before it reaches a
junction is equal to the amperage observed at the point where the branches of the circuit
meet. Kirchhoff’s loop rule, on the other hand, states that the sum of the changes in potential
throughout the whole circuit should be zero.
3. Experimental Design
Control Group: the voltage of the battery
Independent Variables: the magnitude of the resistors used
Dependent Variables: the current and the voltage found on different points of the circuit
4. Procedures
Setup:
a. Set up the PASCO interface. Connect the voltage sensor to (ChA) and current sensor
(ChB) into the interface.
b. Plug the alligator clips on the wires, and connect them to the battery terminals.
c. Set up the circuit, but leave the wire disconnected to the “-“ terminal of the battery.
Procedures: Voltage Sensor
a. Connect the wire lead to the spring clip at the bottom of the battery holder to complete
the circuit.
b. Select the “Monitor Data” tab on the Experimental Menu.
c. Touch the tips of the voltage sensor at the ends of the first resistor. Repeat the process for
each resistor.
d. Measure the voltage throughout the whole circuit.
e. Disconnect the wire at the bottom of the battery holder. Stop DataStudio from recording
any further data.
Procedures: Current Sensor
a. Connect the wire lead to the spring clip at the bottom of the battery holder to complete
the circuit.
b. Select the “Monitor Data” tab on the Experimental Menu
c. Observe the magnitude of current in the Digits display of DataStudio and record the
value.
d. Repeat the process and record the value of the current through each resistor.
e. Measure the current through the entire circuit by placing the current sensors on points A
and B.
f. Stop DataStudio, and disconnect the wire lead from the battery holder.
5. Data
Resistance (Ω)
Voltage (V)
R1 = 10 Ω
V1 = 0.294 V
R2 = 3.3 Ω
V2 = 0.292 V
R3 = 30 Ω
V3 = 0.997 V
R4 = 3.3 Ω
V4 = 0.297 V
R5 = 10 Ω
V5 = 0.294 V
RTotal = 56.6 Ω
VA-B = 1.588 V
Current (theoretical) = 0.028 A
Percent Difference = 125%
Current (A)
I1 = 0.033 A
I2 = 0.031 A
I3 = 0.50 A
I4 = 0.033 A
I5 = 0.033 A
ITotal = 0.63 A
6. Questions
a. Which resistors are in parallel? What do you observe about the voltages of resistors in
parallel?
Resistors 1 and 5 and Resistors 2 and 4 are parallel to one another. The voltages for
resistors 1 and 5 are exactly equal, while the voltages between resistor 2 is 0.005 less than
resistor 4.
b. What is the total of supplied voltage of this circuit? How does this compare to the voltage
drop across the resistors?
The total supplied voltage for the entire circuit was 1.606 V. The total supplied voltage is
greater than the drop in voltage across the resistors. The voltage dropped by a magnitude
of 1.588 V, leaving 0.018 V flowing through the circuit.
c. How does the voltage across R1 and R2 compare to the voltage across R4 and R5? How do
you account for this result?
The voltage across R1 and R2 is higher compared to the voltage across R4 and R5 because
the current has not met any resistance when it arrived at this point. On the other hand,
once the current reaches R4 and R5, it has already met with the R1, R2, and R3 resistors,
severely reducing its voltage.
d. What is the value of the currents going into and out of junction A?
The current moving into and out of junction A is 0.63 A.
e. What is the value of the current going into and out of junction B?
The current moving into and out of junction B is 0.66 A.
f. How many times greater is the value of R1 versus R2? How does this affect the amount of
current that flows through each resistor?
The magnitude of R1 is about 3 times greater than the resistance of R2, meaning that a
greater voltage is prevented by R1 from passing through in comparison to R2. This can be
proven through Ohm’s Law. Dividing 1.606 V by 10 Ω will yield 0.1606 A of current,
while dividing 1.606 V by 3.3 Ω will yield 0.4867 Ω.
7. Conclusion
In this lab, the group coordinated by assigning one person to handle the operation
of the computer, while the other two members performed the experiment. The
investigation started by determining the voltage of the battery being used. Group
members also acquired the resistors needed to satisfy the procedures of the lab. The
group acquired two 3.3 Ω resistors, two 10 Ω resistors, and a 30 Ω resistor (the group
could not find a 27 Ω resistor). It was necessary to determine whether the circuit being
created was either a series or a parallel circuit, and it was found that a parallel circuit is
being created, a circuit appropriate for testing Kirchhoff’s junction rule. Once the proper
set up of the circuit has been achieved, the voltage sensors (with alligator clips at both
ends), were used to determine the voltage at each point of the circuit in the areas were
different resistors were found. The same step was taken with the current sensors. The
information gathered was then used to validate the ideas of Kirchhoff’s rules.
The investigation validated Kirchhoff’s junction rule and loop rule. According to
the junction rule, the magnitude of the current entering the junction will equal the current
leaving the junction. In the experiment, the current moving into the junction A was
measured to be 0.63 A, while the one at junction B was said to be 0.66 A. The
discrepancy may have been due to experimental error, making the difference in values
acceptable. The investigation also validated Kirchhoff’s loop rule. The voltage measured
coming from the battery was 1.606 V, while the total voltage measured from the resistors
was 1.588 V. Another discrepancy was found here due to experimental error, making the
data gathered credible, and therefore validating the loop rule.
Experimental errors can be used to account for the discrepancies of the acquired
data. Some of the resistors may not have been attached properly, consequently making
contact with other wires and metals of the circuit, and therefore transferring current and
voltage. It must also be noted that the circuit is not an ideal system. Resistance from the
wires cannot be assumed to be negligible. This means that a lower magnitude of voltage
and current can be expected in a more realistic situation.