Download PHY104 Lab 7: Kirchoff`s Rules

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PHY104 Lab 7: Kirchoff’s Rules
©1991-2009, James J. DeHaven, Ph.D.
In the preceding labs, you were introduced to the basics of electrical circuitry, simple series and parallel
resistance circuits, as well as the behavior of circuits containing resistors and capacitors.
The simple methods we used to analyze series and parallel resistances break down when we attempt to
apply them to more complex circuits. For more complex circuits, a set of procedures, known as Kirchoff’s
Rules, are employed.
In today’s lab you will apply apply Kirchoff’s Rules to the analysis of several circuits:
1) A simple circuit which can also be analyzed using the methods learned in previous labs
2) More complex circuits
3) Measurement of unknown resistance
4) A practical method for measuring force
Introduction
Series and Parallel Resistances
The two basic configurations for resistors are called “series” and “parallel” and these are shown in a
circuit with a battery in figure 1. Current flows through series resistors sequentially, whereas it “divides up” and
flows “side-by-side” through parallel resistors, much like the flow of a river divides up when the channel breaks
into two or more parts (e.g the Niagara River breaks up into two parts when it flows around Grand Island).
Series resistors act in a circuit as though they were a single lumped equivalent resistance, Req , which is
calculated by simply adding up the values of the various resistors (see figure 2).
As can be seen in figure 2, the basic idea for series resistors is to first calculate the equivalent resistance
by adding up the values of the series resistors. If the battery voltage is known, then the current can be calculated
using Ohm’s Law. Note also that, for resistors in series, the current--all the current--flows through each resistor.
This means that at any point in the circuit, the current in equals the current out.
Once the calculation has been performed it is possible to calculate ΔV, the voltage drop, across one or
more of the resistors, again using Ohm’s law. In figure 2, since all the resistors are the same, the voltage drop is
the same across each resistor, 3.33 volts in this case. Note that the sum of all the voltage drops adds up to the
voltage of the battery. As is shown below in figure 3, the voltage drop across an individual resistor is given
simply by its resistance times the current flowing through it.
--
4.4 kΩ
4.4 kΩ
4.4 kΩ
4.4 kΩ
•
4.4 kΩ
•
4.4 kΩ
10 Volts
Resistors in Series
10 Volts
Resistors in Parallel
Figure 1: Examples of series and parallel resistances
This concept is a striking illustration of the relative nature of the electrical potential (voltage). It must be
described with reference to something. Usually, in a circuit, one of the terminals of the power supply is set at
zero volts, and all voltages are measured relative to it. Referring again to figure 3, if we attach a voltmeter at V1
and Vref, we read the expected voltage of 10 volts.
On the other hand, if we attach the voltmeter to the circuit at V1 and V2, we measure ΔV1, the voltage
drop across R1. If we attach the meter at V1 and V3, we get the combined voltage drop ΔV12 = ΔV1 +ΔV2
across both resistors.
As a result of this analysis, it is evident that as we go around the circuit the the sum of all the ΔV’s
equals zero. For example the voltage drop across all the resistors (a negative number = -10 Volts) is balanced by
the voltage increase across the battery (+10 Volts). We can do this for any point. For example, if we start at V2 ,
V3 - V2 = -3.33 Volts, Vref - V3 = -3.33 Volts, VB - Vref = +10 Volts, V2 - VB = -3.33 Volts. This adds up to
zero volts.
These conclusions may seem trivial, namely that the current into any point on a circuit equals the current
out of it, and that the voltage drops around a circuit add up to zero, but these are, in essence, restatements of
Kirchoff’s rules.
--
Figure 2: Calculations of equivalent resistance, current and voltage for series resistances
--
V4
V3
ΔV3 = IR3
V ref
V2
ΔV2 = I R2
V1
Δ V1 = I R1
VB
Figure 3: Calculation of voltage drops in a voltage divider circuit
That the current into a point must equal the current out of it is a natural result of the conservation of
charge. If some point on a wire does not acquire a static charge (and it doesn’t), then, in any given time interval,
the amount of charge entering it must equal the amount of charge leaving it.
That the potential changes around a circuit must sum to zero is again a natural result of the conservation
of energy. The potential energy of an electron in a circuit is a function only of the location of that electron, and
consequently, all the various increases and decreases that an electron goes through as it moves around a circuit
must add up to zero.
These principles hold in circuits that contain multiple current loops and are generalized as Krchoff’s
rules:
1) At any junction point, the sum of all currents entering the junction must equal the sum of all currents
leaving the junction.
2) The algebraic sum of the changes in potential around any closed path of the circuit must be zero.
We will now apply these rules to the simple parallel circuit we examined last lab (see figure 4).
--
I
I
a
R1 =
4.4 kΩ
R2 =
4.4 kΩ
b
1
I
•
4
c
2
R3=
e
I
d
4.4 kΩ
•
f
3
g
h
10 Volts
Figure 4: Parallel Resistor Circuit Analyzed according to Kirchoff’s Rules
We begin by applying Kirchoff’s current rule to the junction point labeled “c”:
[1] I
4
= I
1
+ I2+I
3
Kirrchoff’s rule for electric potential can be applied to several loops in the circuit giving the following
relations:
[2]
V gh = V cd = V ab = V ef = 10V
Note that our sign convention means that, for example, Vef is the change in potential encountered in
moving from point e to point f. Thus:
V fe = –Vef = –10V
[3] Applying Ohm’s law to [2], we get:
[4] I 1 R1 = I 2 R 2 = I
3
R3
-
We solve for I1 and I3 in terms of I2:
[5]
[6] I
1
= I
2
R2
R1
I3=I
2
R2
R3
Substitute [5] into [1]:
I
4
=I
2
R2 + I + I
2
R1
R2
2
R3
Do a little algebra:
[7] I
4
(
= I 2 R2
1 + 1 + 1
R1
R2
R3
)
But I2R2 is just Vcd, which is the same as the voltage from the battery as a result of [2]:
[8] I
4
= V gh
(
1 + 1 + 1
R1
R2
R3
)
Thus, according to Kirchoff’s Rules, a circuit composed of only parallel resistors will present to a
battery a resistance which has an apparent resistance, whose reciprocal is the sum of the reciprocals of the
parallel resistances. This is precisely what we asserted and experimentally confirmed in last week’s lab:
1 = 1 + 1
+ 1
R eq
R1
R2
R3
R eq =
1467Ω
I =
V
=
R eq
10 Volts
1467Ω
=
1
4.4 kΩ
+
1
4.4 kΩ
= 6.82 x 10 –3 Amps
+
1
4.4 kΩ
-
We can use equation 7 to calculate I2 from I4, and we can easily do the same for I1 and I3:
I1 = I2 = I3 =
–3
1
–3
(6.82 x 10 Amps) = 2.27 x 10 Amps
3
and we recover the unsurprising result that the voltage is the same across all resistors:
V1 = I 1 R 1 = (2.27 x 10 –3 Amps)(4400Ω) = 10 Volts
Why bother with all this nonsense? Aside from the slight aesthetic pleasure of demonstrating that the
rules for calculating equivalent resistances for series and parallel circuits, what is the point? Wouldn’t it be
easier to simply reduce complex circuits to simple ones by viewing them as combinations of series and parallel
circuits? Consider the following circuit:
R1
R3
•
10 Volts
R2
•
R4
Figure 5: Combined series and parallel circuit
To keep the arithmetic simple, lets make all the resistors 1000W. We can systematically go through this
circuit, using our parallel and series rules, successively simplifying things until we represent the entire resistor
network as a single equivalent resistance. We start by lumping together R1 and R2, labeling the equivalent
resistor as R12:
--
1 = 1 + 1
R2
R 12
R1
[9]
1 =
1
1
+
R 12
1000Ω
1000Ω
R 12 = 500Ω
We can replace R1 and R2 in the circuit diagram with an equivalent resistor of 500W, labeled as R12:
R3
R 12
R4
10 Volts
Figure 6: Parallel resistors in figure 5 have been reduced to equivalent resistance
We then combine R12 and R3 using the series resistor rule:
R 123 = R 12 + R 3
[10]
R 123 = 500Ω + 1000Ω
R 123 = 1500Ω
This simplifies the circuit still further:
--
R 123
R4
10 Volts
Figure 7: Now we lump the series resistors
We now have 2 resistors in parallel and can once again utilize our rule for parallel resistors:
1
= 1 + 1
R 1234
R 123 R 4
[11]
1
1
1
=
+
R 1234
1500Ω
1000Ω
R 1234 = 600Ω
This yields a simple circuit containing a single equivalent resistance and a battery:
R 123
10 Volts
Figure 8: And finally the last 2 parallel resistors
-
We can now calculate the current in the circuit, and then backtrack through our earlier diagrams, using
Ohm’s law to calculate the currents and voltages associated with each resistor as well as the power dissipated in
each of them. The process is a bit tedious, but it is straightforward and logical. So who needs Kirchoff’s rules
anyway?
The problem is that not all circuits can be broken down into series and parallel subunits. The circuit
shown below is an example:
R 1 = 2Ω
12 Volts
R 3= 6Ω
6 Volts
R 2 = 4Ω
Figure 9: This circuit cries out for Kirchoff’s rules
To analyze this circuit we must apply Kirchoff’s rules, which we repeat here:
1) At any junction point, the sum of all currents entering the junction must equal the sum of all currents
leaving the junction.
2) The algebraic sum of the changes in potential around any closed path of the circuit must be zero.
A junction point (sometimes referred to as a node) is a point at which three or more conductors are
joined. A closed path (sometimes referred to as a loop) is any path that returns to the same point. We can
identify two loops in our circuit:
--
R 1 = 2Ω
12 Volts
Loop 1
Loop 2
R 3= 6Ω
6 Volts
R 2 = 4Ω
Figure 10: Kirchoff rule loops in circuit to be analyzed
These are of course in addition to the obvious loop around the periphery of the circuit. To use Kirchoff’s
rules we identify the junctions and loops in the circuit and then generate a sufficient number of independent
equations to solve for the unknowns. We need to solve for three currents, and so we will need so generate three
equations. These currents are the ones associated with junction point “b” in the circuit as shown below:
a
R 1 = 2Ω I
1
I
•b
I
2
3
12 Volts
V2
c
V1
R 3= 6Ω
6 Volts
R 2 = 4Ω
•
d
Figure 11: Kirchoff rule junctions in circuit to be analyzed
--
Note that our choice of direction of current in and out of the junction is arbitrary. We have no a priori
knowledge of which direction the current flows and so we simply arbitrarily assume a direction. The signature
of our answers (as we shall see) will inform us about whether or not our original choices of direction were
correct. Note also that if we know the current at “b” we also know it at “d” and so points “b” and “d” provide
redundant information and therefore will not give rise to separate linearly independent equations. Two other
equations will have to be generated (from loops 1 and 2). In fact, Kirchoff rule analysis always generates just
the right number of independent equations to calculate all unknown currents.
We first write down the equation for the currents at the junction point following the prescription that
current in = current out (K’s law number 1):
[12]
I
1
= I
2
+ I
3
To apply Kirchoff’s voltage loop rule we have to adopt a few conventions:
1) We assign a positive voltage when we proceed from the negative to positive plate of the battery;
2) We arbitrarily assign a current direction through each resistor;
3) When we go through a resistor in the direction of the current, there is a voltage drop, i.e. the voltage
is negative;
4) When we go through a resistor and the current is running opposite to the direction we are moving, the
voltage is positive;
5) Once we assign a current to a resistor, we keep that definition, even if it seems counterintuitive.
In the figure below, we have assigned the assumed directions of current flow to our resistors, and
represented this designation by arrows. Do not worry if there seem to be some inconsistencies in current
direction in the drawing. The math will resolve these discrepancies.
--
a
R 1 = 2Ω I
1
I
•b
I
2
3
12 Volts
V2
c
V1
R 3= 6Ω
6 Volts
R 2 = 4Ω
•
d
Figure 12: Assumed current flow in Kirchoff’s rule analysis
We now sum the voltages around the left hand loop (loop 1), going clockwise around the loop:
[13]
V1 – I 1 R 1 – V2 – I 2 R 2 = 0
Note how the current moves in the same direction we are moving as we proceed clockwise, resulting in
negative voltages across the resistors. Note how we go through battery 2 from the positive to the negative plate,
and that this makes V2 negative. We can simplify this equation by plugging in our real values of voltage and
resistance:
[14]
6 volts – I 1 (2Ω) – 12 volts – I 2 (4Ω) = 0
I 1+ 2 I
2
= –3
We can do the same thing in loop 2 (the right hand loop), again proceeding clockwise:
[15]
V2 – I 3 R 3 + I 2 R 2 = 0
-
Note how we have gone through battery 2 in a “positive” direction, making V2 positive. There is a
voltage drop across R3, but the voltage across R2 is now positive because our clockwise motion is now opposite
to the current flow.
Again we can simplify matters by plugging in our actual values of resistance:
[16]
12 volts – I 3 (6Ω) + I 2(4Ω) = 0
3I
3
– 2I 2= 6
Finally we are there. We now have three equations in three unknowns (the three unknown currents):
I
[17]
= I
1
I 1+ 2 I
3I
3
2
2
+ I
3
= –3
– 2I 2= 6
These equations are easily solved simultaneously yielding the following results:
3 Amps
I1= –
11
[18]
15
I 2 = – 11 Amps
12 Amps
I3=
11
Not to worry about the negative signs for the first 2 currents. As we have been saying all along, our
choice of current directions was arbitrary, and it all would come out in the math. The negative signs simply
indicate that we chose wrong for I1 and I2. The magnitudes are correct, but the direction of current flow in R1
and R2 is simply opposite to that implied by the arrows in our drawing (figure 12).
-The Wheatstone Bridge
The Wheatstone Bridge is a very sensitive and useful method for detecting unknown resistances. A
straightforward extension of its principles into AC analysis permits its use in ascertaining unknown AC
impedences, including the effects of capacitance and inductance. The behavior of the bridge can be understood
via a fairly straightforward (but tedious) application of Kirchoff’s rules.
A circuit diagram of a Wheatstone Bridge is shown in figure 13:
R1
V
R2
R5
R3
R4
Figure 13: Circuit Diagram for a Wheatstone Bridge
In practical use, R4 is generally an unknown resistor. R3 is a variable precision resistor, and R1 and R2
are resistors or resistor decade boxes whose resistances are known. The current in R5 is monitored either with a
galvanometer or by measuring the voltage drop across R5 with a voltmeter.
The expressions for the current in R5, as well as the currents in the other resistors and in the battery
circuit, can be derived using Kirchoff’s Rules, and methods for doing so are discussed extensively and in detail
in the Apeendices (A&B). For now, we will simply cite the relevant results.
The expression for current in R5 is:
[19]
I5 =
V (R1R4 − R2R3 )
Δ
-where V is the voltage of the battery and Δ is a 6x6 determinant that we could evaluate if we wanted to or
needed to, but we don’t, at least right nows ,so we won’t; but you can if you’d like to; see Appendix B. Now for
the reason that we don’t need to worry about Δ.
The way we use a Wheatstone Bridge is to “balance the bridge”, that is to say, variable resistor R3 is
adjusted until I5 = 0. The bridge is then said to be in balance. Equation [19] takes on a very simple form in this
case:
[20]
[21]
R1 R4 = R2 R3
or
R1 R3
=
R2 R4
Thus, if you know the values for R1 , R2 , and R3 , you can calculate R4 .
-Experimental
1) Calculate the three currents shown in the circuit below. Then build the circuit, verifying (it is hoped) your
calculations. Use an ammeter to measure the current. Please be careful about hooking up your multimeter as an
ammeter. It is very easy to blow a fuse, or worse!
You can use whatever resistors you want, but they must all have different values of resistance and their values
should all be within an order of magnitude of one another. Use the DC power supply as your battery.
R1
I
I
1
2
R3
•
I
3
R4
V1
R2
•
Figure 14: Circuit Diagram for 1st Kirchoff Problem
Hint: This circuit is very similar to the one analyzed in the introduction. It is missing a battery (which
makes the calculation easier), and has an extra resistor. However you can simplify the right hand loop, making
the algebra much easier.
-2) Repeat what you did in part 1 for the circuit below. Calculate the three currents shown in the circuit. Then
build the circuit, verifying (it is hoped) your calculations. Use an ammeter to measure the current.
You can use whatever resistors you want, but they must all have different values of resistance and their values
should all be within an order of magnitude of one another. Use a DC power supply as your battery.
R1
I
I
1
2
•
I
3
R2
R3
V1
•
Figure 15: Circuit Diagram for 2nd Kirchoff Problem
-3) Build a Wheatstone bridge. Use it to demonstrate that it is possible to measure an unknown resistance. To get
your feet wet, try the following resistances:
Resistor
Value
R1
2200
R2
47
R3
decade box
R4
1K
R5
10, 47K
Table 1: Suggested Resistances for Wheatstone “warm-up”
You can try several methods for detecting I5. One is to connect a voltage meter; a pocket multimeter will
do, as will an oscilloscope or a desktop digital multimeter, although you will probably get the best results from
the latter. Connect it to measure the voltage across R5. When the voltage is zero, the current is zero, and the
bridge is in balance. Try both the 10Ω and the 47kΩ resistors and see which works better.
R1
V
R2
R5
M
R3
Figure 16: Wiring a meter or oscilloscope to detect I5
R4
-You can also use a galvanometer to detect balance on the bridge. The wiring diagram is in figure 17.
R1
R2
S1
V
R5
G
R3
R4
Figure 17: Wiring a galvanometer to detect I5
In figure 17, the resistor, R5, is shown wired in series with a galvanometer, G, and a momentary “tap
switch” labeled S1. The galvanometer does not measure current. Instead, it informs you of the presence of
current in the circuit into which it is wired, deflecting one way or another, depending on the direction of flow of
the current. Again, try both the10W and the 47kW resistors for R5 and see which one works better.
In all your measurements, use a multimeter to obtain “exact” values for your working resistances. You
don’t have to do this for the decade box however
When you have finished investigating an “unknown” resistor, whose resistance you actually know, get a
real unknown resistor from your instructor and see if you can obtain its resistance using the Wheatstone Bridge.
No fair measuring the unknown with a multimeter until after you have found its resistance using the bridge.
4) In the final portion of this experiment, you will use a wheatstone bridge to monitor the change in resistance in
a strain gauge, demonstrating how Kirchoff’s laws can be used to evaluate the output of a simple force
transducer. The theory behind strain gauges is fairly straightforward, and you can read about it starting on page
301 of Brophy. You might also be interested in visiting the Omega Engineering web site to see what they have
to say.
In strain gauge circuits, it is customary to give all the arms approximately the same resistance. The
bridge is nearly in balance under these circumstances and a trim pot (a small variable resistor) can be used to
bring it into balance. When the gauge is strained, its resistance changes, and this changes the potential across
-R5. Usually you can just hook a voltage detector across where R5 usually is and the input resistor of your
detector will serve as R5.
Figure 18 illustrates how you will implement these ideas in your strain gauge circuit.
R1
R2
100Ω
R5
V
SG2
O
DEC
SG1
R3
R4
Figure 18: Using a Wheatstone Bridge with Strain Gauges
You will actually use 2 strain gauges to form 2 arms of the bridge, R2 and R4, as shown in figure 18.
These gauges will be attached to opposite surfaces of a steel bar so that they oppose each other: when the
resistance of one increases, the resistance of the other decreases. This has two salutory effects: the signal is
bigger, and the sign of the signal changes depending on the direction in which the bar is flexed.
Referring to figure 18 again, we see that the decade resistance box will serve as the trimmer, and it is
still located at R3. The other arm of the bridge (R1) will be occuppied by a 100W resistor. As noted above, the
input resistor of the oscilloscope will serve as R5.
Attach the bar so that the gauges form arms R2 and R4. Be careful of the strain gauge leads: they are
very delicate. Bring the bridge as close as possible into balance using the decade box. Make sure that the
bandwidth limiter is on in the scope channel you are using and that the scope is DC coupled. You might also
find that wiring a large capacitor across the scope leads (100mF or more) might help diminish the input noise.
You should be able to use 15 volts to power the bridge without melting anything, and this will also help your
signal
-
Being super careful about not dislodging or shorting your leads, flex the bar (you can use the openings
under the table as fulcrums). You should get a reasonable signal when you do this. In practice, these gauges are
generally connected to differential amplifiers which enhances the S/N and the sensitivity substantially over what
you will see today. However, if you were careful about setting things up you will have successfully
implemented a practical application of Kirchoff’s rules and constructed a force transducer.