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El 04- Electronics 4
Laboratory Exercise 8 – Series Circuits, Part 2
OBJECTIVE: To examine Kirchoff’s Voltage Law which states that the algebraic sum
of all the voltages in a single current loop add up to 0.
MATERIALS: Protoboard; power supply; multimeter; resistors: 2.7K, 3.3K, 4.7K, 6.8K,
10K.
DISCUSSION: In addition to the rules we saw in the previous laboratory exercise, there
is one rule that describes the way voltages appear in a series circuit. Kirchoff’s voltage
law states that the algebraic sum of all the voltages in a single current loop, that is, a
series circuit, should be 0. To put it another way, all the voltages across the various
resistors of a series circuit should add up to the applied voltage. We can see how this all
works using Figure 1:
If we follow the flow of electrons from any arbitrary point in the circuit, and add up the
voltages we encounter as we go, the sum of these voltages should add up to 0 when we
return to the starting point. The voltages must keep polarity in mind. Thus, depending on
how we go around the circuit (as current flows or as opposed to current flow), the
polarities will appear as either a positive change as we pass over a resistor if we are
following electron flow, or a negative change if we are going in the other direction. As
we pass a power supply, we find a change in polarity in the negative direction. If we add
all the values up with polarity considered, the algebraic sum is 0.
The easiest way to approach this is to first follow electron flow around the circuit. For
each resistor, the point where the electrons enter a resistor is more negative than the other
end where the electrons leave the resistor. This would give us a positive polarity to the
voltage across the device. In the case of the battery, the point where the electrons enter is
the most positive point in the circuit, so the point where the electrons leave the battery
must be more negative – thus, the battery exhibits a negative polarity for its voltage.
Consider Figure 1 and the starting point marked A. If we follow electron flow from this
point (or any arbitrary point on the circuit), we see that the first thing we encounter is a
voltage across the bottom resistor. We enter that resistor at it’s negative end and exit at
it’s positive end, and according to the schematic, we have a voltage across that resistor of
30 volts. So, this must be a positive 30 volts algebraically, or +30V. Next, we follow the
electrons around to the right-most resistor, and again enter its most negative point. We
exit its most positive point, and see that as we passed through the resistor, we have a
voltage of 20V that appears across the device. This must also be a positive value
algebraically, so now we have +30V plus +20V so far.
Continuing around the loop counter-clockwise, we enter the most negative end of the top
resistor and exit its positive end. Again, this is an increase in voltage in a positive
direction, and so we now have +30V plus +20V plus +10V. This amounts to +60V so far.
Now, when we encounter the power supply, we enter its most positive end and exit at its
most negative end. Algebraically, this amounts to a negative polarity for its voltage. We
then return to the point A where we started. We now have +30V plus +20V plus +10V
plus –60V. This adds up to 0. Or, we could say that +30V plus +20V plus +10V = +60V.
Note that when we move the 60V to the other side of the equals sign, we must change its
sign.
In this exercise, we will build some circuits and make some measurements and
calculations that will verify these rules for us.
PROCEDURE:
1. Obtain the necessary parts and equipment, and arrange these conveniently on the
workbench.
2. Connect the circuit of Figure 2 below:
3. Apply power to the circuit. Measure the voltage drops across each of the resistors,
and fill in the chart below:
Device:
V Measured:
R1
R2
R3
R4
R5
4. Now, add up the voltages from the chart. Total:
_____________________________________
Compare this to the applied voltage. How do they compare?
___________________________
__________________________________________________________________
___________
5. Dismantle the circuit of Figure 2 and connect the circuit of Figure 3:
6. Apply power. Measure the voltage across R1:
_________________________________
7. Measure the voltage across R2: ____________________________________
8. Describe how you would determine the voltage drop across R3 without measuring
the voltage:
__________________________________________________________________
_________
__________________________________________________________________
_________
9. What is the voltage across R3 by the method of step 8?
______________________________
10. Measure the voltage across R3: _______________________________
11. How do the voltages of steps 9 and 10 compare? Was your explanation in step 8
correct?
__________________________________________________________________
_________
__________________________________________________________________
_________
12. Dismantle the circuit and return the equipment and supplies to their proper
locations. Then, answer the following questions for credit.
REVIEW QUESTIONS:
1. You are given the voltages that appear across four resistors in a series circuit.
Write a formula that calculates the applied voltage.
2. Mark the polarity of the voltages across the resistors of Figure 2.
3. Given a series circuit of three resistors with 50 volts applied. The voltages across
the first two resistors are 40V and 10V. What can we say about the third resistor?
4. Given a series circuit of three resistors with 50 volts applied. The voltages across
the resistors are 0V, 0V, and 50V. What do we know about this circuit?
5. Given a series circuit of three resistors with 50 volts applied. The three resistors
are of equal value. The voltages across these resistors measure 16.6V, 16.6V, and
16.6V. What do we know about this circuit?