Phys 232 – Lab 7 Ch 20
Simple DC and RC Circuits
Equipment: power supply, banana cables, circuit board, switch, 220, 270, 460, & 320 , 1k,two multi-meters, differential voltage
probe, Phys 232 experiment kits: batteries and holders, alligator-clip wires, bulbs and holders, and 1F capacitor.
Objectives
In this lab you will qualitatively and quantitatively demonstrate:
 Ohm’s Law relating voltage, resistance, and current
 The Junction rule relating all current flowing into and out of a junction in a circuit
 The Loop rule relating the voltage drop around a closed loop in a circuit
 The capacitor’s behavior sinking and sourcing charge – facilitating current flow
Overview
Chapters 19 and 20 present microscopic and macroscopic perspectives on circuitry. Following
their example, this lab asks you to construct basic circuits using batteries and light bulbs so you
can see the effects of varying voltage, current and resistance. However, you’ll also use a power
supply, resistors, and digital multi-meters to get more quantitative.
Instrumentation
A digital multi-meter is the go-to tool for evaluating circuitry. It can measure resistances,
currents, and voltages (both steady – DC, and oscillating – AC). In this lab, you will use two
multi-meters like those pictured below (using two rather than just one is merely a convenience;
either could do the other’s job.) You will use the black one to measure current and the other
(yellow) one to measure resistance and electric potential differences (voltage). Be careful to
watch the units because they may adjust automatically as the readings change.
3.14 mV
V~
OFF
V
500
200
500
200
BATT
9V
1.5V
20
2000m
2
200m
20
2000k
200k
20k
2000
200
mA
200

44.4 mA
AC V
DC V
mA/k
1. Ohmmeter: To measure resistance, turn the yellow multimeter’s dial to “2000” in the  range (lower left). Only use
the meter in this mode to find the resistance of an individual
resistor, not on a circuit with current running through it. The
probes from the multi-meter should be touched to either
side of the resistor while the resistor is not in parallel with
any other circuit element(s).
k
DCmA
COM
DCV
2. Voltmeter: Turn the yellow multi-meter’s dial to 20 in the
V range (upper left). The probes from the multi-meter
should be touched to the two points between which you
want to know the potential difference. A positive voltage
reading means that the red probe is at a higher voltage than
the black probe is; a negative reading means the reverse.
3. Ammeter: Turn the dial to “DCmA” and plug a wire into
the “mA/k” socket. The ammeter must be inserted into a
circuit so the current flows through it. A positive current
reading means that the current is flowing into the wire
connected to the “DCmA” socket and a negative reading
means it is flowing in the opposite direction.
Ohm’s Law
I.
Theory
As Chapter 20 discusses, for “Ohmic resistors”,
V= -IR,
captures the interplay between the voltage drop (V) across a resistor, the resistance (R) across
it, and the current flowing through it. The negative sign (neglected or avoided in most texts)
reminds us that the direction of current flow is the direction of the voltage drop, just as the
direction of water flow is downhill. Resistors are specific electrical devices that have been
optimized to obey this relation. For something like a light bulb, this isn’t strictly accurate (or,
rather, the “resistance” is itself a function of the current), but it remains qualitatively true that
increasing the voltage will increase the current.
You can deduce the qualitative strength of current through a bulb from its brightness since a
bulb’s brightness reflects the rate with which it radiates energy which must be the rate with
which energy is brought into the bulb by the electrons that speed through it. As you learned in
Chapter 20, the rate with which energy is transferred to the electrons speeding through a
potential difference, i.e. the power, is
P = IV.
Combining that with Ohm’s law one way or another tells us that
P = I2R=V2/R.
So, for constant current, the brighter the bulb, the more resistive it is, or for constant voltage,
the brighter the bulb, the less resistive it is.
II.
Qualitative Experiment
In the Phys 232 experiment box, you’ll find two D batteries and their
holders, colored wires with alligator clips on the ends (they look like the
name suggests), and little light bulb holders and two kinds of bulbs –
round and oblong.
With one and then the other light bulb, wire-up the simple circuit that’s
illustrated. Note that in the two scenarios (with round or oblong bulb)
you are providing the same voltage difference. So comparing their
brightness, which bulb has the greatest resistance?
According to Ohm’s law, if you decrease the voltage applied across the
bulb, you’d expect less current to be driven through it and thus the bulb
to be dimmer. To see this, move one of the wires so only one battery is
across the bulb.
III.
Quantitative Experiment
You’ll compare the voltage drops across four resistors to the values you’d expect according to
Ohm’s law.
There are four resistors held in place by four springs (each of which is wired to a banana plug at
the end of the board). The resistors are distinguishable by their color stripes: A) red-red-purple,
B) red-purple-brown, C) yellow-purple-brown, and D) orange-orange-purple.
Phys 232
Lab 7
3
1. Set the yellow multi-meter for use as an ohmmeter and measure the resistance of each of
the four resistors on the board. Enter your measurements in the first column of the table
below.
Measured
Resistance
()
Measured
Current
(mA)
Measured
Voltage
(V)
Expected Voltage
(from Ohm’s law)
(V)
A) rd-rd-pr
B) rd-pr-br
C) yl-pr-br
D) or-or-pr
2. To measure the current through and voltage across the first resistor, plug the unconnected
red and black wires (from the power supply and the black multi-meter) into the first two
banana plugs, turn on the black multi-meter (switch on the side) and the power supply.
Schematic
44.4 mA
AC V
DC V
k
DCmA
COM
mA/k
Am p s
DCV
+
5V
-
3.14 mV
OFF
V
500
200
~
Vo lt s
BATT
9V
20
1.5V
2000m
2
200m
20
2000k
200k
20k
2000

V
500
200
R
200
mA
200
3. Record the current through the resistor and voltage across it in the first row of the table.
Just record the magnitudes, don’t worry about the signs. Use Ohm’s law to calculate the
expected voltage from the measured resistance and current.
4. For each of the other resistors, turn off the power supply, and move each wire over just one
plug on the board to connect to the next resistor, and then turn the supply back on. Again,
measure the current and voltage, and calculate the expected voltages using the measured
currents and resistances.
Without getting into the uncertainties of multi-meter’s measurements (often some percent of
the measured value plus a percent of the scale the meter’s set to plus 1 in the last digit shown),
it suffices to say that the voltages you directly measure and those you calculate using Ohm’s law
should be within five to ten percent of each other.
Phys 232
Lab 7
4
Loop Rule: Voltages and Resistors in Series
I.
Theory
Kirchhoff’s Loop Rule: The sum of the voltage steps around any closed loop in a circuit must
equal zero. This follows logically from the fact that the electric potential energy for a charged
particle at a given location is the same regardless of the path the particle took to get there. So if
it travels around any closed loop back to its starting point, its net change in electric potential
energy must be zero; divide by the object’s charge, and you have that the corresponding change
in electric potential must also be zero.
Resistors in Series: Two resistors are in series if they have the same current through them. One
consequence of the loop rule is that the voltage drop across a whole series of resistors is simply
the sum of the voltage drops across each individual one. Putting these two ideas together with
Ohm’s law,
Vser  V1  V2  
Vser   IR1   IR2  
Vser   I R1  R2  
So, for the given voltage, you’d get the same current flowing if you had just one resistor with
resistance
Rser  R1  R2 
II.
Qualitative Experiment
Two light bulbs in series, as illustrated, would present the batteries with
a resistance equal to the sum of their resistances. Should more or less
current flow through the two bulbs rather than when there was just one
bulb?
Consider how that would affect the bulbs’ brightness. Wire up the
circuit and see. Are the bulbs brighter or dimmer than if one or the
other were individually wired up to the batteries?
III.
Quantitative Experiment
1. With the power supply and resistors, set up the circuit shown below by plugging the wire
from the supply into the 4th port and the wire from the black multi-meter into the 1st. You
will measure the voltages between the points shown to test Kirchhoff’s Loop Rule.
RA
2
+
5V
3
RB
1
RC
4
Phys 232
Lab 7
5
2. Measure the voltages between each pair of points (plugs, where the black plug is number 1)
with a voltmeter by touching the black lead of the yellow multi-meter to the first point listed
and the red lead to the second point. Record your measurements below being sure to
include the signs.
3. The potential differences should add up to zero for a round trip around the circuit:
V11  V12  V23  V34  V41 =0. Put another way, the three negative
voltages should add up to be opposite the one positive voltage. Check that is the case
(within a few percent.)
4. Based on the individual resistances that you’d previously measured, what is the sum of the
three resistances used, and thus the single equivalent resistance for the circuit?
5. If that is truly the equivalent resistance, then according to Ohm’s law, it should equal the
power supply’s voltage divided by the current that’s flowing through the circuit. According
to the black multi-meter, what is that current?
6. What then is the ratio of the supply’s voltage to that current?
It should be within a few percent of the equivalent resistance you determined above.
Junction Rule: Currents and Resistors in Parallel
I.
Theory
Kirchhoff’s Junction Rule: The sum of all currents entering a junction must equal the sum of all
currents leaving that junction. This is simply a consequence of conservation of charge in the
same way that ‘conservation of cars’ would dictate that (in steady state) the rate with which
cars enter an intersection must equal the rate with which they leave the intersection. So, if you
have a power supply providing Isup current which splits up through umpteen parallel paths,
I sup  I1  I 2  
Combining this with Ohm’s law and the loop rule (which tells us that the full voltage drop of the
supply is applied across each parallel path individually), yields
I sup 
 Vsup
R1

 Vsup
R2
 1

1
 
   Vsup  
 R1 R2

So, just as much current would be drawn from the supply if, instead of having the umpteen
parallel resistances there were just one path with resistance
1
1
1



Rpar R1 R2
Phys 232
II.
Lab 7
6
Qualitative Experiment
To a pretty good approximation, the batteries maintain a
constant voltage regardless of the load (unless the load
has extremely low resistance, and so draws a lot of
current.) So whether you attach one light bulb or two in
parallel shouldn’t significantly change the voltage that the
batteries maintain.
Wire up the illustrated circuit. Connect and disconnect
one of the bulbs and observe the brightness of the other
– it shouldn’t (hardly) change if the battery’s voltage
across and thus current through the bulb is unchanged.
(In actuality, you may notice that the other bulb very
slightly dims; se the text’s section 20.4 for a discussion of
batteries and their effective internal resistance.)
III.
Quantitative Experiment
1. Set up the circuit that’s illustrated schematically to the left below and is illustrated a little
more realistically to the right below. You’ve probably already noticed the four little ‘jumper’
wires connecting six springs on the circuit board; these will help. Move resistor A and
resistor B to complete the circuit. The sharp ‘ice pick’ can be used to wedge apart rungs of a
spring to make room for inserting a resistor’s leg. Plug the two cables into the last and
third-from-last red ports (furthest from the black one) to connect the power supply to the
bottom two springs.
Sup
junction
B
junction
A
Sup
+
5V
B
A
RA
RB
-
To test Kirchhoff’s Junction Rule, you will measure the currents into and out of the junction
indicated with a dot.
2. Dial the yellow multi-meter to the 200 mA position so it can measure current, then insert its
two probe tips where the “Sup” wire’s ends are and remove the Sup wire. Turn on the
power supply, and the yellow multi-meter will measure the current provided by the supply
into the junction.
3. Replace the Sup wire, and, in turn, do the same with the A wire to measure the current
passing out of the junction to resistor A, and then the B wire to measure the current passing
to resistor B.
Phys 232
Lab 7
7
According to the junction rule, the total current flowing out (IA + IB) should equal the current
flowing in (Isup). Your measured values may not be perfectly equal, but they should be
within a few percent of each other.
Capacitors: Charging and Discharging
I.
Theory
Consider the circuit illustrated below. It is similar to that discussed in section 20.6 of your text.
The switch initially connects the capacitor across the power supply and leaves the resistor
dangling so the capacitor gets charged up. Then the switch is flipped to connect the capacitor
across the resistor and leave the supply dangling.
switch
+
-
Vsup
+
C
R
-
Assuming negligible electric field in the connecting wires and corresponding voltage drop along
the wires, the loop rule (from conservation of energy) tells us:
Vcapacitor  Vresistor  0 .
Since Vcapacitor  Qc C and Vresistor   IR , we can write this as

Qc C  IR  0 .
Where I  dQR dt is the rate with which charge flows through the resistor. Rather subtly,
while the capacitor is discharging, every morsel of charge that flows through the resistor is a
morsel that flows off the capacitor, thus reducing the capacitor’s charge: dQc  dQR . So the
current through the wire can be rephrased as I   dQc dt . Thus, the relation can be written
as
dQc
  dQc 
1
  RC
Qc .
Qc C  
 R  0 or
dt
dt


The latter form says ‘the rate with which the charge on the capacitor decreases is proportional
to the amount of charge on the capacitor.’ This is a differential equation describing the
decrease in charge on the capacitor as a function of time. Later in your career you’ll get familiar
with solving such equations, but for now it suffices to say that this particular equation is solved
by Qc  Qo e
t
 RC
(plug this in for Qc in the equation and see that you get a true statement).
Phys 232
Lab 7
8
Since the voltage drop across the capacitor is proportional to the charge on the plates,
Vc  Qc / C , we similarly have
Vc  V0 e
t
 RC
,
where V0 is the initial voltage difference across the capacitor at time t = 0. That would be the
supply’s voltage if t = 0 just when the switch is flipped.
The combination RC is the ‘exponential time constant’ for the circuit (Ohm*farad = second).
Qualitatively, the larger it is, the longer it takes for the capacitor to discharge.
II.
Qualitative Experiment
As a capacitor discharges across a light bulb, the voltage across capacitor and bulb exponentially
decays, and so must the current and radiated power, thus brightness diminishes. How long that
takes depends on the capacitor-resistor’s RC time constant, and thus scales with the bulb’s
resistance.
1. Wire up one of the circuits illustrated using the round light bulb and paying special attention
to the capacitor’s orientation – the long leg must be connected to the + terminal.
+
+
+
+
2. Switch between the two circuits illustrated. When you do so, the bulb will immediately
brighten and then gradually dim. Roughly time that dimming (you don’t need to be terribly
precise – counting in your head will suffice.) This will give you a ballpark for the RC time
constant of the bulb – capacitor combination.
3. Now repeat but using the oblong bulb.
With which bulb does the capacitor take longer to discharge?
Check that that’s consistent with your very fist answer in this lab (to the question of which
bulb is more resistive).
Phys 232
Lab 7
9
III.
Quantitative Experiment
You’re going to charge a 1-farad capacitor with the batteries and then discharge it across a 1k
resistor. You’ll use LoggerPro to monitor the voltage for 10 minutes.
1. In the light bulb – capacitor circuit you just explored, set it up once more to charge the
capacitor (as illustrated on the left).
2. After the capacitor is charged (the bulb’s dimed out), replace the bulb with the 1k resistor
(brown – black – red).
3. Plug a “differential Voltage Probe” into LabPro; Clip the black lead of the probe to the
negative leg of the capacitor (on the side of the capacitor marked with a white stripe and
with the short leg) and the red lead to the other leg of the capacitor.
4. Open the “Capacitor” file from the Physics Experiments/Phys 232 folder.
5. Switch the circuit to the other configuration so the capacitor begins discharging, and
immediately hit the green “collect” button.
6. Once the data has been collected, you can tell Logger Pro to do an exponential curve fit. To
do that, select “Curve fit…” from the Analyze menu, and define an exponential function of
the form A*exp(-t/B); ask instructor for help.
What’s the exponential decay time? (note: since the time axis in minutes, you’ll need to
multiply B by 60 to get it in seconds.)
7. Disassemble the circuit and use the multimeter to measure the resistance of the “1-k”
resistor.
Since the decay time should equal RC, and you have values for that time and for R, what
must be the value of C?
That should be within about 10% of the 1 farad printed on it.