Kirchhoff’s Laws
General Physics II
Kirchhoff’s Laws
General Physics II
Here, we will explore Kirchhoff’s Laws and see how to apply them in order to completely describe a complicated
circuit.
Kirchhoff ’s Laws
There are two laws that we can use to evaluate any multiloop circuit:
Current Law: The total current going into a node is equal to the total current leaving the node.
As an example, for the node pictured below, Kirchhoff’s Current Law allows us to write I1 + I3 = I2 + I4 .
I4
I1
I2
I3
Voltage Law: The sum of the voltage increases and decreases around a circuit will be 0.
This is the harder of the two laws, and we will explore it in the following examples.
Problem
Compute the current in each branch of the below circuit.
Property of Garrett Higginbotham.
Please send any suggestions to ghiggie@uab.edu
1
Last Modified: June 24, 2015
Kirchhoff’s Laws
General Physics II
Solution
These problems can be very confusing. Thus, we will develop a set of guidelines to help us with any multi-loop
problem.
First, at every battery, draw the current in that branch moving from the negative terminal of the battery to the
positive terminal. If there is a branch without a battery, then the current in that branch can be drawn in either
direction. It is advisable to draw the current completely going along that branch. This gives us the following image:
Next, we need to choose a direction (within each loop) to measure in. I like to do it from the negative terminal
to the positive terminal of the battery, going in that direction around the loop. Thus, I would go clockwise around
the top loop. But this poses a problem with the bottom loop, which has two batteries pointed in different directions.
Property of Garrett Higginbotham.
Please send any suggestions to ghiggie@uab.edu
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Last Modified: June 24, 2015
Kirchhoff’s Laws
General Physics II
In a situation like this, you can choose your direction arbitrarily. For the sake of example (it isn’t the most efficient
way), I will measure counter-clockwise from the top battery around the lower loop. When I do this, I like to draw
circular arrows within the loop to indicate my directions. This gives the following diagram:
OK, so now we’re ready to actually start solving the problem. We first want to use Kirchhoff’s Current Law.
You can choose either node to measure from, so I will measure from node F . Notice that the currents I1 and I2 are
going into the node, and I3 is leaving the node. Thus, we have
I1 + I2 = I3 .
If you instead measured from node C, you would obtain the equation I3 = I1 + I2 , by similar reasoning. Now, let’s
try to apply Kirchhoff’s Voltage Law.
Let’s first focus on the loop ABCF . Moving clockwise from the battery, we first encounter a voltage of 8 V . Then
we encounter a resistor, which decreases our voltage by 2I1 . Finally, we encounter another resistor, which further
decreases our voltage by 4I3 . Putting all of this together, we obtain the equation
8 − 2I1 − 4I3 = 0.
Now, let’s look at loop CF ED. Measuring from the top battery, we first encounter a voltage of 8 V . Moving
counter-clockwise around the loop, we then encounter a resistor which decreases our voltage by 2I1 . We then
encounter another resistor, but there’s an issue this time: the current is going against us this time. When this
happens, just use the negative of that current. Thus, the voltage decrease due to this resistor is 2(−I2 ). Finally,
Property of Garrett Higginbotham.
Please send any suggestions to ghiggie@uab.edu
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Last Modified: June 24, 2015
Kirchhoff’s Laws
General Physics II
we encounter another battery, but this one is pointed in the wrong direction. Like with the resistor, just take the
negative of this voltage. Thus, this battery contributes −6 V to our equation. Putting all of this together, we obtain
the equation
8 − 2I1 − 2(−I2 ) + (−6) = 0 ⇒ 2 − 2I1 + 2I2 = 0.
To reiterate, we now have the following equations:
I1 + I2 = I3
(1)
8 − 2I1 − 4I3 = 0
(2)
2 − 2I1 + 2I2 = 0
(3)
Substituting equation (1) into equation (2) and simplifying, we get 3I1 + 2I2 = 4. Rearranging equation (3),
we arrive at 2I1 − 2I2 = 2. Adding these two equations together and solving for I1 , we obtain I1 = 1.2 A .
Substituting this into equation (3), we obtain I2 = 0.2 A . Finally, substituting this values into equation (1), we
obtain I3 = 1.4 A .
If you follow the general guidelines discussed in this example, you should do fine on any multi-loop circuit problem.
Property of Garrett Higginbotham.
Please send any suggestions to ghiggie@uab.edu
4
Last Modified: June 24, 2015