Physics 151
Complex Circuits
Postlab Assignment
Name __________________________________
In Complex Circuits you continued your investigation of electrical circuits. Throughout the lab you used
the dc power supply to power your circuits, the DMM configured as a voltmeter to measure potential
differences, and the DMM configured as an ammeter to measure currents. Were these three devices
behaving as ideal power supplies, ideal voltmeters and ideal ammeters? What does ideal even mean?
No electronic device is ideal. In this postlab you will explore how the behavior of real power supplies,
voltmeters and ammeters deviates from the ideal, and how this “non-idealness” is modeled.
DC Power Supply
Think about the power supply you have been using in lab. It has a red (+) terminal and a black (-) terminal
to which you connect wires. Its function is to maintain a potential difference between those two terminals.
For example, if you dial-in 5 V the power supply will keep the red terminal 5 V higher in potential than the
black terminal.
DC Power Supply
Red (+) terminal
E = 5V
Rload
Black (-) terminal
Figure 1: Ideal power supply
Ideally, we envision the power supply casing housing an ideal source of EMF, set at 5 V in this example, as
shown in Figure 1 above. If it really is ideal, that means it will maintain the 5 V potential difference
between the red and black terminals no matter what we connect to those terminals. For example, if we
connect a 100 Ω resistor, called the load resistor, it will have to deliver 5 V/100 Ω = 0.05 A of current if it
is to maintain a 5 V potential difference across the load. If instead we connect a 10 Ω resistor to the power
supply, it will have to deliver 5 V/10 Ω = 0.5 A of current if it is to maintain a 5 V potential difference
across the load. If we connect a 1 Ω resistor to the power supply, it will have to deliver 5 V/1 Ω = 5 A of
current if it is to maintain a 5 V potential difference across the load. Notice that as the resistance connected
to the supply gets smaller the amount of current the supply must deliver gets larger. However, a real power
supply cannot deliver an arbitrarily large amount of current, so we would expect deviations from ideal
behavior especially as the load resistance becomes too small.
A real power supply is modeled as an ideal source of EMF in series with, what is called, the power supply’s
internal resistance. In our minds, as shown in Figure 2, we envision the actual casing of the power supply
housing this ideal source of EMF in series with its internal resistance, which we’ll call Rsupply.
1
DC Power Supply
Red (+) terminal
Rsupply
Rload
E = 5V
Black (-) terminal
Figure 2: Model of a real power supply
How does the internal resistance alter the behavior of an ideal source of EMF? To investigate, let’s
continue with the example of setting the power supply to provide a 5 V potential difference across a load.
With your knowledge of series circuits you can see that not all of the 5 V will appear across the load. Some
fraction of it will appear across the internal resistance of the supply. Furthermore, if a very low resistance
is connected to the supply, the current cannot rise arbitrarily high, the current is limited by the internal
resistance of the supply. In fact, with this model, for the limiting case of zero load resistance (shortcircuiting the supply), the current can be no higher than the ideal EMF divided by Rsupply.
Investigate on your own the behavior of a real supply by filling in the table below. The power supply is set
for 5 V and has an internal resistance of 1Ω. Different load resistances are presented and you must
calculate the current in the load and the potential difference that actually appears across the load. (Carry
out your calculations to four significant figures in amps and volts.)
Actual
Current
Ideal
Terminal
(Load)
Voltage
EMF
Rsupply
Rload
Ideal
Current
5.000 V
1.000 Ω
∞ (no load)
0.0000 A
5.000V
5.000 V
1.000 Ω
100.0 Ω
0.0500A
5.000V
5.000 V
1.000 Ω
10.00 Ω
0.5000A
5.000V
5.000 V
1.000 Ω
1.000 Ω
5.000A
5.000V
5.000 V
1.000 Ω
0.000 Ω (short)
∞
5.000V
Actual
Terminal
(Load)
Voltage
Remember, in all these cases, the supply has been set to maintain 5 V across the load. In other words,
before you connect the load, you turn the knob until the meter on the front face of the supply says 5 V, fully
expecting 5 V to appear across the load. But just because you dial-in 5 V doesn’t mean you will get 5 V
when the load is connected. Right? (Rhetorical.)
1.
Based on your calculations should you worry about this non-ideal behavior when the load
resistance is large compared to the supply’s internal resistance or when the load resistance is
comparable to the supply’s internal resistance? (Note: This is not a yes or no question.)
2.
In this respect is a power supply better if it has a large internal resistance or a small internal
resistance?
2
Ammeter
Consider the circuit shown in Figure 3(a). Assuming the power supply to be an ideal source of EMF there
is 0.5 A of current in the circuit.
Ammeter
R
ammeter
E = 5V
Rload
10Ω
Rload
E = 5V
(a)
10Ω
(b)
Figure 3
Let’s say you want to measure the current with an ammeter. You would expect a good ammeter to measure
0.5 A. However, very few measurements, if any, are completely unobtrusive. In other words, every time
you use an instrument to measure something you can expect the instrument to somehow alter the value of
the quantity you are measuring. It is part of your job as a scientist or engineer to be aware of your
instrument’s influence on the system you are investigating, and in many cases, to design your test or
experiment so that this influence is accounted for or made insignificant.
Ammeter’s certainly behave in this less-than-ideal fashion. The very act of placing an ammeter in a circuit,
to measure the current in the circuit, reduces the amount of current in the circuit. This non-ideal behavior
can be modeled by assigning the ammeter an internal resistance of its own that will be called Rammeter.
3.
The actual circuit is shown in Figure 3(a). We fully expect 0.5 A of current in this circuit. The
test circuit is shown in Figure 3(b). The rectangle represents the physical casing of the ammeter.
Note that it has two probes connecting it inline with the loop of the circuit in order to measure the
current in the circuit. The ammeter’s internal resistance is represented by a resistor symbol placed
inside of the ammeter. If the ammeter’s internal resistance is 1 Ω what will the ammeter actually
measure? Show your work.
4.
Based on this calculation is an ammeter better if it has a large internal resistance or a small internal
resistance? Explain.
3
Voltmeter
Consider the circuit shown in Figure 4(a). Assuming the power supply to be an ideal source of EMF there
is 5 V dropped across each of the two identical 800 kΩ resistors.
R1
R1
E
10V
800kΩ
R2
800kΩ
E
10V
Voltmeter
800kΩ
R2
800kΩ
Rvoltmeter
(a)
(b)
Figure 4
Let’s say you want to measure the potential drop across R2 with a voltmeter. You would expect a good
voltmeter to measure 5 V. But just like the ammeter, a real voltmeter will alter the very quantity it is trying
to measure. This non-ideal behavior of the voltmeter can be modeled by assigning the voltmeter its own
internal resistance that will be called Rvoltmeter.
5.
The actual circuit is shown in Figure 4(a). We fully expect 5 V to be dropped across R2. The test
circuit is shown in Figure 4(b). The rectangle represents the physical casing of the voltmeter.
Note that it has two probes connected across R2 in order to measure the potential difference. The
voltmeter’s internal resistance is represented by a resistor symbol placed inside of the voltmeter.
If the voltmeter’s internal resistance is 2 MΩ what will the voltmeter actually measure? Show
your work.
6.
Based on this calculation is a voltmeter better if it has a large internal resistance or a small internal
resistance? Explain.
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