Kirchhoff's Laws

PHYS 1112L/2212L
Kirchhoff’s Laws
Objective
To practice setting up and measuring complex circuits and to practice solving
circuits using Kirchhoff’s Laws.
Theory in a Nutshell
Sometimes you may be faced with complex circuits with multiple loops and
numerous junctions. In order to find the current in various parts of the circuit, you
can use Kirchhoff’s Laws, which can be stated simply this way:
• The amount of current coming into a junction equals the amount of current
going out of it. That is, there are no sources or sinks of current; it does not
magically appear or disappear.
• In going around a closed loop of a circuit, the sum of voltage changes must
equal zero. That is, if you gain any voltage going across some component in a
circuit, that must be offset by an equal loss somewhere else in the loop.
Let us consider an example of how to apply these rules. See the circuit below.
You have a DC power supply and a parallel combination of resistors R 1 and R 2 ,
with currents through them equal to I 1 and I 2 , respectively. Total current I comes
out of the power supply on the left and flows clockwise until it reaches the
junction at the top of the circuit. We can visualize this circuit as follows:
The total current I comes in from the left and splits into two branches, I 1 and I 2 .
Kirchhoff’s junction rule tells us that I = I 1 + I 2 . You can draw the currents in
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either direction. Be bold! The worst that can happen (assuming you do your
calculations correctly) is that you will get a negative number for your current.
That just means it goes opposite the direction you drew it. Take a deep breath,
decide which direction each current must go, and write your equations for each
junction.
Now consider the closed loop consisting of the power supply and R 1 . We can
think of it this way:
Remember this is just one loop of the circuit. We drew the current going
clockwise. Start at the lower left corner (you can start anywhere, but lower left
seems like a convenient place to start).
We cross the power supply from negative to positive (remember the longer line
denotes the positive side of the DC supply). That means we add voltage in the
amount the power supply is putting out. So we have +V. Now we continue on
around the circuit, going through only wire until we reach R 1 . While we are just
going through wire, these are ideal wires, so there is no resistance and no loss of
voltage. So now we get to R 1 . Since we are going in the direction of the current,
there is a voltage drop across the resistor, given by Ohm’s law as the product of
the current and the resistance. So we have –I 1 R 1 for the change in voltage.
Now we are at the bottom of the loop and there is just more wire back to the
starting place, so no gain or loss of voltage. We can write Kirchhoff’s loop rule
for this loop as follows: +V – I 1 R 1 = 0.
If we went the other way around the circuit, the signs would change. If we went
from positive to negative across the DC supply, the voltage change would be –V.
If we went across the resistor opposite to the direction of the current, the voltage
change would be +I 1 R 1 . In either case, you wind up with V = I 1 R 1 . You can
repeat this process for the outside loop of the circuit.
When solving these circuits, decide which direction you believe the currents go,
and draw them. Then write the equation for each junction, where the current
coming in equals the current coming out. Then look at each loop and write an
equation for the loop rule. That should give you enough equations to solve even
the most complex circuit. And of course, be careful with your algebra.
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Worked Example (review thoroughly before coming to lab)
Let’s work an example with numbers. Consider the circuit below. The emf is 6
volts. R 1 = 100 ohms, and R 2 = R 3 = 50 ohms. What is the current through each
resistor
We notice that the current that flows through the emf (let’s call it I 1 ) flows around
to the junction at a, and then splits into two parts (call them I 2 and I 3 ). They flow
down through their respective resistors and then rejoin at junction b. The
combined current (I 1 ) flows through R 1 and then back through the emf.
From the junction rule, you can write I 1 = I 2 + I 3 . That’s one equation. We need
two more.
Consider the loop that runs through the emf, around through a, down through R 2 ,
through b, then through R 1 . The sum of voltage gains and drops must equal zero.
If we begin at the lower left corner of the circuit, and work clockwise, first we
pass through the emf from negative to positive. This adds 6 V. Then we move
along until we get to R 2 . We drew the current I 2 going down, so the voltage will
drop by an amount equal to I 2 R 2 . Now we move through R 1 , and again, since we
drew the current going to the left, the voltage will drop by I 1 R 1 . So the equation
reads:
+6 V – I 2 R 2 – I 1 R 1 = 0
That’s two equations. We need one more.
Consider the loop that runs through the emf, around through a, down through R 3 ,
through b, then through R 1 . The sum of voltage gains and drops must equal zero.
If we begin at the lower left corner of the circuit, and work clockwise, first we
pass through the emf from negative to positive. This adds 6 V. Then we move
along until we get to R 3 . We drew the current I 3 going down, so the voltage will
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drop by an amount equal to I 3 R 3 . Now we move through R 1 , and again, since we
drew the current going to the left, the voltage will drop by I 1 R 1 . So the equation
reads:
+6 V – I 3 R 3 – I 1 R 1 = 0
That’s three equations in three unknowns. You can solve it any way that makes
sense to you, but here’s one way. Subtract the two loop equations.
+6 V – I 2 R 2 – I 1 R 1 = 0
+6 V – I 3 R 3 – I 1 R 1 = 0
0 V - I 2 R 2 – (-I 3 R 3 ) – I 1 R 1 – (-I 1 R 1 ) = 0
-I 2 R 2 + I 3 R 3 = 0 so I 2 R 2 = I 3 R 3
Now, since in this case, R 2 = R 3 , you can also say that I 2 = I 3 .
Back to the junction rule, I 1 = I 2 + I 3 . Since I 2 = I 3 , then I 1 = 2I 2 .
Go back to the first loop equation and rewrite in terms of I 2 .
+6 V – I 2 R 2 – 2I 2 R 1 = 0
Substitute in the values of R 1 and R 2 that you were given in the problem.
+6 – I 2 (50) – 2I 2 (100) = 0
Simplify: +6 – 50I 2 – 200I 2 = +6 – 250I 2 = 0
+6 = 250I 2 I 2 = 6/250, or 0.024 amp = I 3
I 1 = 2I 2 = 0.048 amp
Are those correct How would we check One way is to review the circuit. With
your extensive knowledge of series and parallel circuits, you could calculate the
net resistance in the circuit. 100 ohms in series with a parallel combination of two
50 ohms equals 100 plus 1/(1/50 + 1/50) = 100 + 25 = 125Ω. So the current
flowing through the emf = I 1 = 6V/125Ω = 0.048A. And since R 2 = R 3 , then I 2 =
I 3 , so each must be one-half of 0.048, or 0.024. Bingo.
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You should attempt to solve each of the circuits below before coming to lab class.
Follow the same general procedure:
• Identify all junctions.
• Label all emfs, resistors, and currents. Use subscripts wisely to keep
everything straight.
• Decide and draw how you think the current will flow in each part of the
circuit.
• Write the junction rule equation for each junction.
• Write the loop rule equation for each loop.
• Do the best algebra you have ever done in your life.
Equipment
Power supplies
Resistors
Multimeters
Wires, clips, etc.
Procedure
In this lab you will do three circuits. In each case, before you set up the circuit,
calculate the current in each loop using Kirchhoff’s Laws (this is sometimes called
solving the circuit). Then set up the circuit and measure each current. If you get a
substantial difference, you have an issue either with your calculations or with your
circuit; work at it until you resolve the issue. Show all work to receive credit for
this lab. Note whether currents are positive or negative, and explain why.
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Circuit #1 – one emf
R 1 = 2000 Ω
R 2 = R 4 = 1000 Ω
R 3 = 1500 Ω
Once you have solved the circuit, set it up and measure the current in each loop.
Calculate percent errors for your measured values.
Circuit #2 – two emfs
R 1 = 1000 Ω
R 2 = 2000 Ω
R 3 = 3000 Ω
Once you have solved the circuit, set it up and measure the current in each loop.
Pay particular attention to positive and negative values, and explain each.
Calculate percent errors for your measured values.
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Circuit #3 = three emfs
ε 1 = 8 V
ε 2 = ε 3 = 4 V
R 1 = 1000 Ω
R 2 = 2000 Ω
R 3 = 6000 Ω
Once you have solved the circuit, set it up and measure the current in each loop.
Pay particular attention to positive and negative values, and explain each.
Calculate percent errors for your measured values.
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Questions
1. Kirchhoff’s laws – the loop rule and the junction rule – are each based on a
fundamental conservation principle in physics. For each rule, explain in your own
words what conservation principle is involved and how it leads logically to the
rule.
2. Are your measured currents generally larger or smaller than the theoretical
values you calculated
3. We have done this lab assuming negligible resistance for these three things:
• The power supply
• The ammeter
• The wires
If any or all of these have non-negligible resistance, explain qualitatively
how that might affect your measured currents. Would each of these make
the current larger, or smaller
4. Extra credit: find a documented estimate (such as from the manufacturer)
for the resistances of the three components listed above, and show quantitatively
how large an error, if any, such resistances would cause in your results. Your
instructor will provide you the information on where the components are
purchased if you need that. Show all work.
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