Chapter 18 Circuit Problems
General Procedure
To solve these, we’re going to use Kirchoff’s laws several times. There are two loops, and as such there are
three different currents. I named them IL , IR ,and IM , as the current in the left wire, right wire, and middle
wire. There are three unknown currents, and so we need three equations to be able to solve for them.
A key point in doing these problems is the initial step. You must GUESS the direction of the current that
is flowing in the wire. There is no right or wrong guess, you simply have to assign a direction to each of the
currents, and remain consistent throughout your calculation. At the end of the problem, minus signs will
show up that will correct or affirm your guesses. I usually guess (random assignment of direction)
IL flows clockwise, as in UP the left side.
IM flows DOWN the middle wire.
IR flows counter clockwise, as in UP the right side.
Again, I assert that this is an arbitrary choice. Anderson seems to use down down up. You could have chosen
different directions and gotten the right answer. The correctness of your answer depends only on whether or
not you stayed consistent with your choice. We will generally apply the loop rule to the left and right loops,
and then the junction rule to either one of the two junctions.
Junction rules: After assigning the directions of the three currents, pick one of the two junctions. It
doesn’t matter which one. Now, follow along each of the three arrows and note: If you travel along the
direction of the arrow and come INTO the junction you chose, then this current is going INTO the junction.
If you follow along an arrow and it leads you away from the junction you chose (towards the other junction),
then this current is LEAVING the junction. Two of them will be doing one thing, and the third will be doing
another. Either way, set the total current going IN equal to the total current going out. This is the first of
the three Kirchoff equations that we need.
When going around the loop, pick a starting point (anywhere, I usually pick a corner), and pick a direction,
clockwise or counter clockwise. This direction has nothing to do with the currents, it just marks the way you
are going to add things up. Follow these rules when you encounter the corresponding objects:
Battery rules: Regardless of the direction of the (guessed) current, ADD the voltage if you go from the
negative terminal to the positive terminal. SUBTRACT the voltage if you go from the positive terminal to
the negative terminal. By ’go’ I mean the way you are going around the loop(CW or CCW).
Resistor rules:
This does depend on the direction of our guesses. Remember to stay consistent. ADD the IR term if you
are going opposite the direction you guessed for the current. SUBTRACT the term if you are going in the
same direction of the assumed current.
Problem 17
Let’s apply the rules now. First I’m going to add up the voltages on the leftmost loop. Again I assigned the
left wire to have a current going UP, middle wire DOWN, and right wire UP. Once you finish a loop, set the
sum equal to zero.
Left loop: (I am starting at the bottom left corner, and going around clockwise, down the middle wire
and back to the bottom left corner).
−IL (8Ω) − IM (5Ω) − IM (1Ω) − 4V = 0
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Right loop: (Starting from the bottom right corner, going counterclockwise, down the middle wire and
back to the bottom right corner).
+12V − IR (1Ω) − IR (3Ω) − IM (5Ω) − IM (1Ω) − 4V = 0
The junction rule is easiest. Pick a junction (place where wires meet), and set the currents going in equal
to the currents going out. At the top junction:
IM is leaving the junction, IR and IL are coming into the junction, so:
IR + IL = IM .
So we have three equations and three unknowns. Take the three equations and make them look nicer,
combine like terms, etc:
6IM + 8IL = 4
4IR + 6IM = 12
IR + IL = IM
This becomes an algebra problem with lots of substitution. Be careful not to make any mistakes. Once
you’ve solved for the three currents, note which ones come out negative (if any). These negative signs will
serve to correct our guesses for the direction of the current. If any of the three currents come out negative, then
simply flip the direction that you originally guessed the current to point in (and remove the negative sign),
and that is the direction of the ACTUAL current. Report your answers as a number, and a DIRECTION,
such as clockwise or counterclockwise (not negative or positive).
Problem 21
This problem is slightly different in that now we do know one of the currents, but we don’t know how strong
one of the batteries is. Here they told us the direction and magnitude of IL , namely IL = 2.0A upwards.
We can still apply Kirchoff’s rules as described above. Again, using the same conventions that I used for
problem 17:
Left loop: (I am starting at the bottom left corner, and going around clockwise, down the middle wire
and back to the bottom left corner).
+E − 2(1) − 2(3) − IM (5) − IM (1) − 8 = 0
Right loop: (Starting from the bottom right corner, going counterclockwise, down the middle wire and
back to the bottom right corner).
−IR (12) − IM (5) − IM (1) − 8 = 0
The junction rule is easiest. Pick a junction (place where wires meet), and set the currents going in equal
to the currents going out. At the top junction:
IM is leaving the junction, IR and IL are coming into the junction, so:
IR + IL = IM .
So we have three equations and three unknowns. Take the three equations and make them look nicer,
combine like terms, etc:
6IM + 16 = E
6IM + 12IR = −8
IR + IL = IM
This is again an algebra problem, with unknowns IM , IR , andE. We really are interested in E, so you’ll
want to substitute out the two currents first to save yourself a bit of work. Otherwise, its the same procedure
as before.
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