Physics 309—Intermediate Laboratory Indiana University Department of Physics
Lab #1: Electrical Measurements I—Resistance
Goal: Learn to measure basic electrical quantities; study the effect of measurement
apparatus on the quantities being measured by investigating the internal resistances of
digital multimeters (DMMs) and batteries. Learn the concept of equivalent circuits.
Equipment: Battery, DMMs, assorted resistors and a variable resistor box, Proto-Board,
connectors, jumper wires, … .
1
Internal Resistance of a Voltmeter
When we want to know the voltage between two points, we connect a voltmeter (a DMM
set to a voltage scale, see Figure 1) between the points and read the result.
Images of Typical Digital Multimeters (DMMs)
Select Function (Volts, Ohms,
Amps) with Dial. Hit Range
button repeatedly to select range
(e.g. 10mV, 1V, 10V, 100V).
Make sure not set to autorange.
Select Function (Volts, Ohms,
Amps) and Range (e.g. 10mV,
1V, 10V, 100V) with Dial.
Figure 1: For both types of DMM, make sure you correctly set the Function
dial to DC voltage (symbol 𝑉𝑉� or 𝑉𝑉� ) for measurements of steady voltages and to
AC voltage (symbol 𝑉𝑉� ) for measurements of oscillating voltages.
Always make sure that one of your connections is to Common (also called
Ground) and the other to the connection appropriate for the voltage, current, or
resistance measurements you are making (different on different DMMs).
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An ideal multimeter has infinite internal resistance and has no effect on the voltages in the
circuit we are studying, but any real voltmeter has a finite resistance 𝑅𝑅𝑉𝑉 which changes the
voltage we are trying to measure. A ‘real’ voltmeter always draws some current and thus
represents a (small) load on the circuit. We can think of a real voltmeter as if it were an
equivalent circuit with finite resistance 𝑅𝑅𝑉𝑉 in parallel with an ideal voltmeter, as shown in
Figure 2.
In this exercise we will find this internal resistance so you can estimate when it might have
a significant effect on the voltage measurements you will make later in the course. In order
to measure the internal resistance 𝑅𝑅𝑉𝑉 of your voltmeter, set up a circuit with two DMMs as
shown in Figures 2 and 3. Why is 𝑉𝑉1 not equal to 𝑉𝑉2? This circuit is a voltage divider; with
𝑅𝑅𝑉𝑉
series resistor 𝑅𝑅𝑆𝑆 . Show that 𝑉𝑉2 = 𝑅𝑅 +𝑅𝑅
𝑉𝑉1. Solve for 𝑅𝑅𝑉𝑉 in terms of 𝑉𝑉1, 𝑉𝑉2 and 𝑅𝑅𝑆𝑆
𝑆𝑆
𝑉𝑉
The 0-15V power supply of the Proto-Board provides a variable voltage. Use a 10MΩ (1 ×
107 ohms), ¼ watt resistor for the series resistor 𝑅𝑅𝑆𝑆 . Measure 𝑅𝑅𝑆𝑆 with an ohmmeter (the
DMM on the Ω scale) before building the circuit, since the actual resistance of a resistor
may differ significantly from its nominal value. Your fingers are conductive, and can affect
your resistance measurement, so keep them away from any metal parts in the circuit while
measuring 𝑅𝑅𝑆𝑆 . Vary the supply voltage between 0V and 15V in steps of about 3V (to get a
voltage of 0 just turn off the power supply), and calculate the internal (input) resistance of
DMM2 from these data using your calculation above (Use a fixed voltage range for DMM2
to make this measurement. Do not use autorange.).
DMM2
V2
V1
R 10M
S
ideal voltmeter
R infinite
DMM1
V2
0-15 volts
R
s
v
V =0
0
Figure 2: Circuit for measuring the internal resistance 𝑅𝑅𝑉𝑉 of voltmeter DMM2.
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Figure 3: Photo of the setup for the internal resistance measurement in part 1 and Figure
2. You can easily remove the large breadboard to see how the nodes and busses connect
(see also Figure 4).
2
Internal Resistance of an Ampère Meter
To measure the current that flows through an electrical circuit, you can interrupt the circuit
Figure 4: Details of Breadboard on Protoboard. Left, Top View. Right, Wire
connections between holes in board. Note that the holes at the top and bottom are
connected together with long horizontal wires. The holes in the middle are connected
in groups of five by short vertical wires.
and insert an Ampère meter or Ammeter (a DMM set to a Current range.) so the current
flows through the Ammeter. The resistance of an ideal ammeter is zero and does not change
the current. However, a real ammeter has a small internal resistance 𝑅𝑅𝐴𝐴 , as in Figure 5. To
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measure 𝑅𝑅𝐴𝐴 , set up the circuit shown in Figure 5. Use two 1 kΩ ¼ watt resistors in parallel
for 𝑅𝑅𝑆𝑆 . Make sure you measure 𝑅𝑅𝑆𝑆 before completing the circuit.
Use the 15 volt variable supply in the Proto Board to vary the current into DMM2.
Configure DMM2 as a current meter with range 0-20 mA and be sure to use the correct
plugs on DMM2. Use DMM1 to measure the voltage across the real ammeter DMM2.
Make sure you have DMM1 connected to measure 𝑽𝑽𝟏𝟏 , not 𝑽𝑽𝑺𝑺 . The connection is not
the same as that in Part 1. If you measure 𝑉𝑉𝑆𝑆 rather than 𝑉𝑉1, you can still derive 𝑅𝑅𝐴𝐴 , but
your calculation will be different and your result less accurate. Assuming DMM2
𝑉𝑉
accurately reports the current, calculate 𝑅𝑅𝐴𝐴 = 𝐼𝐼1 , where I is the current through the
ammeter. Note that by measuring 𝑉𝑉1 and I you do not need to know the value of 𝑅𝑅𝑆𝑆 . If you
measured 𝑉𝑉𝑆𝑆 and I, you would need to know 𝑅𝑅𝑆𝑆 very accurately to derive 𝑅𝑅𝐴𝐴 .
The resistors will get warm when they dissipate a power near or above their rating of ¼W.
Assuming that 𝑉𝑉1 ≈ 0, calculate the voltage 𝑉𝑉𝑆𝑆 where the power dissipation in each resistor
exceeds ¼ watt (remember 𝑃𝑃 = 𝐼𝐼 2 𝑅𝑅). You can touch the resistors to sense this effect (be
careful not to burn yourself).
DMM2 (current meter)
V
1
1k ea.
R
DMM1
V
R
A
S
S
0-15 volts
A
V =0
0
ideal current meter
R=0
Figure 5: Circuit for measuring the internal resistance 𝑅𝑅𝐴𝐴 of ammeter DMM2.
3
Batteries as a Voltage Source, Internal Resistance of a Battery
A battery is a device that produces a voltage difference between its poles. When we connect
a load resistor 𝑅𝑅𝐿𝐿 between these poles, a current 𝐼𝐼 flows through the load resistor. As we
decrease RL, the current through the resistor increases. However, the battery cannot supply
very large currents, so for small 𝑅𝑅𝐿𝐿 , the voltage 𝑉𝑉1 across the battery terminals drops. That
is, the battery behaves like an equivalent circuit with an output resistance 𝑅𝑅𝑆𝑆 in series with
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an ideal voltage supply. This output resistance limits the current flow the battery can
produce.
In order to measure 𝑅𝑅𝑆𝑆 , connect a variable load 𝑅𝑅𝐿𝐿 (resistor box) and a voltmeter across the
battery as shown in Figure 6. You will obtain better results using a fresh battery. Use the
same DMM2 and settings as in the previous part of the experiment to measure the current
𝐼𝐼 through the load resistor. If 𝑅𝑅𝐿𝐿 is very large, 𝐼𝐼 ≈ 0 and 𝑉𝑉1 = 𝑉𝑉𝑆𝑆 . Assuming that the
𝑉𝑉 (𝑅𝑅 +𝑅𝑅 )
internal resistance of the voltmeter DMM1 is infinite, show that 𝑉𝑉1 = 𝑅𝑅 𝑆𝑆+ 𝑅𝑅𝐿𝐿 + 𝐴𝐴𝑅𝑅 . where RA
𝐿𝐿
is the value you measured in part 2.
𝐴𝐴
𝑆𝑆
Measure V1 and I for different values of 𝑅𝑅𝐿𝐿 . Start with 𝑅𝑅𝐿𝐿 ≈ 1000Ω and decrease the value
of 𝑅𝑅𝐿𝐿 until the current increases to 𝐼𝐼 ≲ 30mA. Using a very small RL causes the battery to
produce a very large current, which can damage the battery (we have plenty so don’t worry
too much). Use values of RL which produce 𝐼𝐼 > 30mA only briefly (for the duration of
your measurement). Measure the actual values of 𝑅𝑅𝐿𝐿 that you used after you complete your
measurements of 𝑉𝑉1 and 𝐼𝐼. Measure 𝑉𝑉𝑆𝑆 before and after the experiment and use the average
of the two values in your derivations. Make sure you include the difference between the
before and after values of 𝑉𝑉𝑆𝑆 as a source of error in your error analysis.
You can derive the value of 𝑅𝑅𝑆𝑆 in two different ways in this experiment. First consider the
current in the entire circuit 𝐼𝐼. The voltage across DMM1 is 𝑉𝑉𝑆𝑆 minus the voltage across 𝑅𝑅𝑆𝑆 ,
𝑉𝑉1 = 𝑉𝑉𝑆𝑆 − 𝑅𝑅𝑆𝑆 𝐼𝐼.
So:
𝑉𝑉𝑆𝑆 − 𝑉𝑉1
𝑅𝑅𝑆𝑆 =
.
(eq. 1)
𝐼𝐼
In this case, you don’t need to know 𝑅𝑅𝐴𝐴 or 𝑅𝑅𝐿𝐿 to measure 𝑅𝑅𝑆𝑆 .
Alternatively, you know that:
Substituting for 𝐼𝐼,
𝐼𝐼(𝑅𝑅𝐴𝐴 + 𝑅𝑅𝐿𝐿 ) = 𝑉𝑉1 , 𝐼𝐼 =
𝑉𝑉1
.
𝑅𝑅𝐴𝐴 + 𝑅𝑅𝐿𝐿
𝑉𝑉𝑆𝑆 − 𝑉𝑉1
(𝑅𝑅𝐴𝐴 + 𝑅𝑅𝐿𝐿 ).
(eq. 2)
𝑉𝑉1
In this case, we don’t need to know 𝐼𝐼, which means we could have simplified the circuit by
𝑉𝑉 −𝑉𝑉
getting rid of the ammeter entirely (in which case the result would be 𝑅𝑅𝑆𝑆 = 𝑆𝑆 1 𝑅𝑅𝐿𝐿 ).
𝑉𝑉
𝑅𝑅𝑆𝑆 =
1
Calculate 𝑅𝑅𝑆𝑆 using both equation 1 and equation 2. Are your results the same? Which is
more accurate? Why?
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Figure 6: Circuit for measuring the internal resistance 𝑅𝑅𝑆𝑆 of a battery.
4
Adjustable Power Supply and Output Resistance of a Voltage Divider
Just as a battery has an internal resistance, a power supply also has an internal resistance,
which means that the actual output voltage will be smaller than the nominal output voltage
for loads with low resistance. We can explore the significance of this effect more easily if
we build a “power supply” with an internal resistance we control. To explore the idea of
an equivalent circuit in more detail, we will explore the use of a voltage divider to obtain
a selected voltage from a fixed-voltage power supply. In Figure 7 we can think of the
voltage across 𝑅𝑅2 as the output of a power supply producing a nominal voltage 𝑉𝑉𝑃𝑃𝑃𝑃 , (the
voltage the equivalent power supply produces with no load (𝑅𝑅𝐿𝐿 = ∞) consisting of the real
power supply plus the voltage divider provided by 𝑅𝑅1 and 𝑅𝑅2 ).
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Calculate
𝑉𝑉2
𝑉𝑉1
as a function of 𝑅𝑅𝐿𝐿 . Your result should show that the power supply plus
divider behaves just like an equivalent power supply composed of an ideal voltage supply
with a series resistance RS, which turns out to be the parallel combination of resistors 𝑅𝑅1
𝑅𝑅 𝑅𝑅
and 𝑅𝑅2 �𝑅𝑅𝑆𝑆 = 𝑅𝑅 1+𝑅𝑅2 �. This result is an example of Thevenin's Theorem for two-terminal
1
2
linear networks. If you are unfamiliar with Thevenin's Theorem, many web resources (such
V
Equivalent Power Supply
1
R
1
V2
DMM1
15 volts
R
DMM2
2
V=0
s
R
L
resistor box
100k ... 500 ohms
Figure 7: Circuit for creating an equivalent power supply with a selected voltage
output using a fixed power supply and two resistors as a voltage divider.
Equivalent Power Supply
Thevenin's theorm
v
Equivalent Power Supply
R
R
1
V
S
equivalent to
0
R
2
𝑉𝑉𝑃𝑃𝑃𝑃
VS
𝑉𝑉𝑃𝑃𝑃𝑃
Figure 8: Thevinin’s theorem for analyzing the circuit in Figure 7 as an equivalent
circuit, a power supply with nominal output voltage 𝑉𝑉𝑃𝑃𝑃𝑃 . The dashed boxes are
equivalent circuits to the dashed box in Figure 7.
as Wikipedia) explain it. If 𝑅𝑅𝐿𝐿 ≫ 𝑅𝑅2 , then 𝑉𝑉2 ≈ 𝑉𝑉𝑃𝑃𝑃𝑃 . When 𝑅𝑅𝐿𝐿 ≈ 𝑅𝑅2 you will find 𝑉𝑉2 <
𝑉𝑉𝑃𝑃𝑃𝑃 . If 𝑅𝑅𝐿𝐿 ≪ 𝑅𝑅2 you will find 𝑉𝑉2 ≪ 𝑉𝑉𝑃𝑃𝑃𝑃 . Why?
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To construct the circuit in Figure 7, use 10𝑘Ω and 5𝑘Ω for 𝑅1 and 𝑅2 . Vary the resistance 𝑅𝐿 ,
using nominal values of 𝑅𝐿 = 0.5, 1.0, 3.3, 6.8, 10, 33, 100kΩ, ∞ (to get 𝑅𝐿 = ∞ just
disconnect one terminal of the load resistor)Measure the actual values of these resistors with an
ohmmeter before constructing the circuit. Record the output voltage for each measured value of
𝑅𝐿 . Determine the expected value of 𝑉𝑆 in terms of 𝑅1 , 𝑅2 and 𝑉1. Calculate Rs from R1 and R2.
Compare your measured voltages, V2, with the expected values for VL (the voltage across 𝑅𝐿 in
the equivalent circuit with 𝑅𝐿 in series with Rs and Vs). How well do your data match the theory?
Questions to discuss in your discussion and conclusion sections:
1) Questions related to part 1. Under what circumstances will the internal resistance of a
voltmeter significantly affect the voltage you are trying to measure? Specifically,
consider a situation in which the load resistance in the circuit between the two points at
which you are measuring is 𝑅1 , the remainder of the circuit has a series resistance 𝑅2 and
the internal resistance of the voltmeter is 𝑅𝑉 . What will be the fractional difference
𝑉𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 −𝑉𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
(
𝑉𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
) in the voltage across 𝑅1 between the unperturbed circuit and the
circuit with the voltmeter in place as a function of 𝑅1 , 𝑅2 and 𝑅𝑉 ? Does this difference
depend on 𝑉𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑 ? Define what you mean by “Significant.” Use your measured
value of 𝑅𝑉 to determine a condition on 𝑅1 and 𝑅2 so that the change in voltage is less
than 5%.
2) Questions related to part 2. Under what circumstances will the internal resistance of an
ammeter significantly affect the current you are trying to measure? Specifically, consider
a situation in which the series resistance in the circuit you are measuring is 𝑅1 and the
internal resistance of the ammeter is 𝑅𝐴 . What will be the fractional difference
𝐼
( 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑
−𝐼𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
𝐼𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
) in the current through the circuit between the unperturbed circuit
and the circuit with the ammeter in place as a function of 𝑅1 and 𝑅𝐴 ? Does this difference
depend on 𝐼𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑 ? Define what you mean by “Significant.” Use your measured
value of 𝑅𝐴 to determine a condition on 𝑅1 so that the change in voltage is less than 5%.
3) Questions related to parts 3 and 4.
a) Under what circumstances will the internal resistance of a battery or power supply
significantly affect the voltage or current you are trying to apply to a circuit?
Specifically, consider a situation in which the load resistance in the circuit you are
measuring is 𝑅𝐿 and the internal resistance of the battery or power supply is 𝑅𝑆 . What
will be the fractional difference (
𝑉𝑎𝑝𝑝𝑙𝑖𝑒𝑑 −𝑉𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
𝑉𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
) in the voltage applied to the
load as a function of 𝑅𝐿 and 𝑅𝑆 ? Does this difference depend on 𝑉𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑 ? Define
what you mean by “Significant.” Use your measured values of 𝑅𝑆 to determine a
condition on 𝑅𝐿 so that the change in voltage is less than 5%.
b) The current is a slightly different calculation. Imagine a perfect power supply which
𝑉
produced a current through your load 𝐼𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑 = 𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
. Let 𝐼𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑 be the
𝑅
𝐿
current through the real circuit. What is the fractional difference in current
𝐼
( 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
𝐼
−𝐼𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑
) as a function of 𝑅𝐿 and 𝑅𝑆 ? Is the dependence of the
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difference on 𝑅𝐿 and 𝑅𝑆 the same for current and voltage? Use your measured values
of 𝑅𝑆 to determine a condition on 𝑅𝐿 so that the change in current is less than 5%.
c) If you are designing a power supply to have a very well defined voltage output,
should RS be large or small? Why? If you are designing a power supply to have a
very well defined current output should RS be large or small? Why? Which is a better
voltage supply, the battery, or the equivalent power supply in npart 4? Why? Which is
a better current supply? Why?
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