1
Name: ________________
Physics 12
Direct Current Circuits Lab
PART 1 – Circuit Elements in Series
In the circuit drawn in Figure 1 at the right the circuit
elements #1, #2, and #3 (in this case we are using light
bulbs) are said to be connected IN SERIES. That is, they are
connected in a series one right after another.
First you are to measure the electric CURRENT flowing in
different parts of the series circuit. Electric Current is the
number of charges per unit time that pass a point in a circuit.
The unit of current that we will use is the AMPERE, or amp for short. If 1 coulomb of charge
passes a point in a circuit in 1 second then 1 amp of current is flowing past that point. Thus by
definition:
1 AMP = 1 COULOMB/SECOND.
The currents you will be measuring in this experiment will be in amps or milliamps (ma), where 1
ma is 1/1000 or 0.001 amp. Connect the ammeter as instructed by Mr. Nagy in the circuit at
the positions listed in Table 1 and record the current.
Table 1.
Position of Ammeter
Current (amps)
Between A & B
Between C & D
Between E & F
Between G & F
Using the data from Table 1 and the definition of current answer the following:
Q-1
What can you conclude about the current anywhere in a series circuit?
________________________________________________________________
Q-2
How many coulombs of charge leave the power supply each second? _____________
Q-3
How many coulombs arrive at the other terminal of the power supply each second?
______________
Q-4
Recall that 1 coulomb of charge is carried by 6.25 x 1018 electrons or protons. Each
proton or electron has 1 elementary charge (1.6 x 10-19 C). How many elementary
charges pass through each light bulb in each second? _______________________
2
Now remove the ammeter from the circuit. As demonstrated by Mr. Nagy, connect the
voltmeter to measure the potential difference (voltage) across each light bulb and across the
power supply. Record your measurements in Table 2.
Table 2.
Position of Voltmeter
Potential Difference (V)
Across #1
Across #2
Across #3
Across the Power Supply
Let’s see what these potential differences mean in a physical sense. First assume that positive
charges are the only charges that flow in the circuit (this is the historical “convention”, and
equivalent arguments can be made for negative charges with the same results). Notice in
Figure 1 that there is a positive and a negative terminal on the power supply. On the positive
terminal there is an excess positive charge. Hence, the positive charges at that terminal
experience a repulsive force due to this excess positive charge. These charges then have
electrical potential energy. When an external circuit is attached to the power supply the
positive charges flow through the circuit to the negative terminal. As they arrive at the
negative terminal, the power supply has the characteristic that it forces these positive
charges to move within the power supply from the negative to the positive terminal thereby
again increasing their electrical potential energy. THE CHANGE IN ELECTRICAL POTENTIAL
ENERGY PER UNIT CHARGE IS CALLED THE POTENTIAL DIFFERENCE OR VOLTAGE
between the terminals.
If the energy is measured in joules and the charge is in coulombs then the potential difference
(PD) has units of joules per coulomb. The joule/coulomb is given the name Volt. That is:
1 VOLT = 1 JOULE/COULOMB
Thus the potential difference across any circuit element is a measure of the difference in the
potential energy per unit charge on one side of the element as opposed to the other side of the
element. For example, if a battery has a PD of 1.5 volts between its terminals, this means that
the battery increases the potential energy of each coulomb of charge that pass through the
battery by 1.5 joules. Using the definition discussed above and the data in Table 2 fill in the
following:
The potential difference across the terminals of the power supply was _________. This means
that each coulomb of charge has its potential energy increased by ____________ as it passes
from the negative to positive terminal within the power supply. This also means that each
individual elementary charge gained _______________ of potential energy as it passed from
the negative to the positive terminal.
3
Now consider what happens to the charges as they flow through the external circuit. Let us
assume that the charges lose no energy in the wires between the light bulbs. However, when
they move through the bulbs there is light and heat energy given off. Therefore, there must
have been a significant amount of the electrical potential energy of the charges converted into
these forms. The potential difference across each bulb tells us how much energy is converted
to these forms per coulomb of charge that passes through the bulb. For example, the potential
difference across bulb #1 was measured to be _________ volts (refer to measurement in
Table 2). This means that each coulomb of charge that passed through bulb #1 gave up _____
joules of potential energy which radiated away in the form of heat and light energy. Each
coulomb also gave up ______ joules of potential energy as it passed through bulb #2 and
_____ joules of potential energy as it passed through bulb #3.
Q-5
What was the total amount of potential energy lost by each coulomb of charge that
passed through the circuit from the positive terminal (point A) of the power supply to
the negative terminal (point H)?
Q-6
How much potential energy was given to each coulomb of charge originally by the power
supply?
Q-7
______________________________
______________________________
How does the potential energy per unit charge supplied by the power supply compare
with the total potential energy per unit charge given up at the charges move through
the light bulb circuit?
______________________________
From the answers to Q-5, 6, & 7 you should recognize the LAW OF CONSERVATION OF
ENERGY.
To summarize what you have stated in Q-7:
THE SUM OF ALL THE POTENTIAL ENERGY INCREASES PER UNIT CHARGE
EQUALS THE SUM OF ALL THE POTENTIAL ENERGY DECREASES PER UNIT
CHARGE AROUND A CLOSED LOOP IN A CIRCUIT.
Or since potential energy increase per unit charge MEANS a voltage increase and
potential energy decrease per unit charge MEANS a voltage decrease, than the above
can be restated as:
THE SUM OF THE VOLTAGE INCREASES EQUALS THE SUM OF ALL THE VOLTAGE
DECREASES AROUND ANY CLOSED LOOP IN A CIRCUIT.
4
POWER
Recall from previous work that POWER IS THE ENERGY TRANSFERED INTO OR OUT OF A
SYSTEM PER UNIT TIME. For example, if a battery supplies 2.5 joules of energy to a circuit
every 5 seconds then it is supplying energy to the circuit at a rate of 2.5 joules/5 seconds =
0.5 joules/sec or 0.5 watts. Remember:
1 WATT = 1 JOULES/SECOND
The battery then is said to be supplying power to the circuit at a rate of 0.5 watts.
Now let’s see how the power supplied or dissipated in a circuit is related to things like current
and potential difference (alias voltage). Suppose there is a current of 0.5 amps flowing through
a light bulb and that the difference in potential across the light bulb is 2.0 volts. This means
that __________ coulombs of charge pass through the bulb during each second and each
coulomb that passes through the bulb loses _________ joules of potential energy. Therefore,
_________ joules of electrical potential energy are being converted into heat and light every
second.
Q-8
Since the power dissipated by the light bulb is the energy transferred to it per unit
time, what is the power dissipated in the above example? ______________________
Note that in the example above the power dissipated can be found by merely multiplying the
current flowing through the bulb by the potential difference across the bulb. Generally then,
the power supplied by or dissipated in any circuit element is found by multiplying the current
through it by the voltage across it. That is:
POWER = CURRENT x VOLTAGE or P=VI
Comparing the units we can see that
AMPS x VOLTS =
× = = WATTS.
Using your data from Tables 1 & 2 calculate the following:
i.
ii.
iii.
iv.
v.
Q-9
The power supplied by the power supply = (_____)x(_____) = ________
The power dissipated in bulb #1 = (_____)x(_____) =
________
The power dissipated in bulb #2 =
________
The power dissipated in bulb #3 =
________
The total power dissipated in bulbs 1, 2, & 3 =
________
What can you conclude from your answers to i and v above?
____________________________________________________________
5
PART 2 – Circuit Elements in Parallel
The set-up for this portion is shown in figure 2 at the
right. As in Part 1 the circuit elements are small light
bulbs. The bulbs marked #1 and #2 are said to be
connected IN PARALLEL since they are connected to
common points at each side. Note that by our previous
definition bulbs #3 & #4 are connected in series with
one another.
Connect the ammeter between points A and B as labeled in the diagram, then between B and C,
then between B and E, then between G and H. These measurements will determine the current
flowing out of the power supply (IAB), the current through #1 (IBC), the current through #2
(IBE), and the current through #3 (IGH). Record your meter readings in Table 3.
Q-10
From what you have learned about currents in series circuits, what do you predict for
the current flowing between points I and J? between K and L?
Predictions: IIJ = _______, IKL = _______
Check your predictions with the ammeter and enter them in Table 3.
Table 3.
Position
Current (amps)
Between A & B
Between B & C
Between B & E
Between G & H
Between I & J
Between K & L
Remove the ammeter from the circuit and with the voltmeter measure the potential
differences ACROSS each of the circuit elements and record them in Table 4.
Table 4.
Position
Potential Difference (V)
Across #1
Across #2
Across #3
Across #4
Across the power supply
Q-11
Referring to your data in Table 3, how does the sum of the currents in the parallel
elements (#1 & #2) compare with the current in the rest of the circuit?
__________________________________________________________
6
Refer to your data in Table 4 and fill in the following:
The potential energy supplied to each coulomb of charge passing through the power supply was
_____________________.
The potential energy lost by each coulomb of charge that passed through bulb #1 was
_______________________.
The potential energy lost by each coulomb of charge that passed through bulb #2 was
____________. *** through bulb #3 was __________. *** through bulb #4 was _________.
Note that the potential energy lost per unit charge as the charges pass through either of the
parallel circuit elements is the same. That is, THE POTENTIAL DIFFERENCE ACROSS
PARALLEL ELEMENTS IN A CIRCUIT IS THE SAME.
Example – Suppose you are given the situation shown in Figure 3 where you are
to measure the currents I1, I2,I3,I4, & I5 and the potential difference across
each circuit element.
Q-12
How would you expect I1 to compare with I5?____________________
Q-13
How would you expect the sum of I2+I3+I4 to compare with I1?
________________________________________________
Q-14
How would you expect the potential difference across each of the
circuit elements to compare?
_____________________________________________
Now refer back to Figure 2 and your data in Tables 3 & 4.
Suppose 1 coulomb of charge moves through the power supply, then through bulb #1, then
through bulb #3, then bulb #4, then back to its starting point. Compare the increases in its
potential energy with the decreases as it moves along this path.
_____________________________________________________________
Suppose another coulomb of charge follows the path through the power supply, but passes
through bulb #2, then through bulb #3, then bulb #4, then back to its starting point. Compare
the increases in its potential energy with the decreases as it moves along this closed path.
_____________________________________________________________
Q-15
Referring to the comparisons above, does the rule stated in Part 1 for series circuits
“THE SUM OF THE VOLTAGE INCREASES EQUALS THE SUM OF ALL THE VOLTAGE
DECREASES AROUND ANY CLOSED LOOP IN A CIRCUIT” hold also for circuits
having elements that are connected in parallel with one another?
___________________________________________________________
7
POWER
Using the same concepts and definitions presented in Part 1 and your data from Tables 3 & 4,
calculate the following:
i.
ii.
iii.
iv.
v.
vi.
Q-16
The power supplied by the power supply = (_____)x(_____) = ________
The power dissipated in bulb #1 =
(_____)x(_____) = ________
The power dissipated in bulb #2 =
__________
The power dissipated in bulb #3 =
__________
The power dissipated in bulb #4 =
__________
The total power dissipated in bulbs 1, 2, 3, & 4 =
__________
What can you conclude from your answers to i. and vi. above?
_______________________________________________________
PART 3 – Ohm’s Law
The set up for part 3 is shown at right in figure 4. The symbol
represents a resistor, A represents an ammeter, and V represents
a voltmeter. In this part of the lab you will vary the voltage across the
power supply and measure the voltage across the resistor and the current
through it.
By varying the voltage across the power supply, the voltage across the
resistor can be varied. Vary the voltage across the resistor from about 0.5
volts to about 6.0 volts and measure the resulting current through the
resistor. Record your measurements in Table 5.
Table 5.
Voltage across R (V)
Current through R (amps)
Graph the voltage across the resistor (vertically) versus the current in amps (horizontally)
8
Does the graph show a relationship between voltage across and current in a resistor?
_______________________________________________________________
Restating your conclusion: the potential difference (or voltage) across a resistor is equal to a
constant of proportionality times the current through the resistor.
Or,
V=IxR,
where R is the constant of proportionality.
The constant R is called the RESISTANCE of the resistor. For the resistor in this part of the
lab the numerical value of R can be found by taking the slope of the V versus I graph.
Q-17
What was the resistance, R, of the resistor used in this part of the experiment? Be
sure to include units from the graph!
R= _________________________
The unit of resistance is the OHM ( Ω ). By definition: 1 OHM = 1 VOLT/AMP.
The equation V = IR is called OHM’ LAW. Assume that the light bulbs used in Parts 1 & 2 obey
Ohm’s Law. Calculate the following resistances:
R of bulb #1 in Part 1 = (_____)÷(_____) = ________
R of bulb #1 in Part 2 = (_____)÷(_____) = ________
R of bulb #2 in Part 2 = (_____)÷(_____) = ________
PART 3 – Ohm’s Law
In Figure 5 at the right the power supply maintains a potential
difference of 6.0 volts between its terminals. The current out
of the power supply is 0.08 amps and R1 = 30 Ω and R2 = 70 Ω.
The potential difference across R3 is 4 volts.
1.
A coulomb of positive charge at point B has less
potential energy than a coulomb of positive charge at
point A. How much less? ___________
2. What is the resistance of R3? _________________
3. What is the potential difference across R1? ___________ ; across R2? ___________
4. What is the current through R1? __________ ; through R2? _______________
5. What is the decrease in potential energy per coulomb of charge that pass from point A
through R1 then through R3 to point B? _____________________________
6. What is the decrease in potential energy per coulomb of charge that pass from point A
through R2 then through R3 to point B? _____________________________
7. What is the power output of the power supply? _______________________
8. How much energy is supplied to the circuit in 5 seconds? ________________
9. What is the total energy dissipated in all three resistors during every 5 second period?
_________________________
9