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Last time we went over Kirchhoff’s Current and VOltage laws. Now we’re going to see
how those laws can be used in conjunction with Ohm’s law to solve a circuit.
Kirchhoff's Current Law or KCL defines a relationship between all the currents in a circuit.
Kirchhoff's Voltage Law or KVL defines a relationship between all the voltages.
Ohm's Law defines the relationship between voltage and current.
So using these three together allows us to determine all the voltages and currents in
a resistive circuit.
Let’s start with the following circuit. For purposes of this problem and I’m going
to define solving the problem as determining all the voltages and currents in the circuit.
That being the case, we will have to begin by defining all the voltages and currents
that are not known. Let’s start with the eleven Ohm resistor. If we define the voltage
across the five Ohm resistor, as V2, then again, the current must enter the positive
terminal of that voltage. The same is true for voltage across the 12 Ohm resistor and
the current through it and the voltage across the eight Ohm resistor and the current through
it. Since the current from the source is also unknown, I can assign the variable to it also.
In this case, because it is a source, the current will leave the positive terminal of
the voltage source.
At this point, we have three laws that apply to circuits, Ohm’s Law, Kirchhoff’s Current
Law and Kirchhoff’s Voltage Law. We’ll begin by writing the Ohm’s Law relationship
for the resistors so we write V1 equals 11 Ohms times I1, V2 equals five Ohms times I2,
V3 equals twelve Ohms times I3 and V4 equals eight Ohms times I4.
Now let’s turn to Kirchhoff’s current law. To apply Kirchhoff’s current law, we
will first identify the nodes at which we’re going to write KCL equations. The circuit
has four nodes. We can write the equations for all the unknown currents if we write equations
for the top three nodes. To install the unknowns are already included in the equations we will
write for these three nodes, writing an equation for the bottom node would provide no new information.
Once the nodes are identified, was helpful to label them for purposes of reference. I’ll
call the left-hand node, node A, the center node, node B and the right-hand node node
C.
We can then write or KCL equations for each node. I’ll write them in the form that the
current entering must equal the currents leaving. So, at node A Is enters and I1 leaves. At
node B, I1 enters and I2 and I3 leave. At node C, I3 enters and I4 leaves.
We can then move on to write KVL equations for the unknown voltages. To apply KVL, we
need to identify loops that cover all unknown voltages. Let’s identify the left-hand one
as L1 and the right-hand loop as L2. As I mentioned in the previous video, if, when
I go around in the direction of the arrow, I go from a negative to a positive I will
call that a voltage increase and write it has a positive voltage in the equation. If
I go from a positive to a negative, I will call that a voltage decrease and write it
as a negative voltage in the equation. So the equation for L1 is: we go up by 15 volts,
down by V1 and down by V2. So we get 15 volts minus V1 minus V2 equals zero, since all the
voltages around any closed loop must equal zero. Going around L2, we go up by V2 down
by V3 and down by V4. So we end up with V2 minus V3 minus V4 equals zero.
Now we have all the equations that describe the currents and voltage relationships for
the circuit. We have for equations from Ohms law. Three equations from KCL, and two equations
from KVL. That gives us nine equations with nine unknowns. The rest is algebra. Let’s
begin by combining the Ohms law equations with a Kirchhoff’s voltage law equations.
We can do this by simply substituting in the Ohms law relationships into the KVL equations.
This results in the following two equations.
Already we have reduced our system of equations to five equations with five unknowns. Since
I as only appears in one equation and is not related to any other quantities and we’ll
ignore it for now. We can then use substitution to reduce our system of equations. We can
substitute equation for I1 into the relationship below and also substitute equation for I3
into the other equation below, resulting in two equations with two unknowns.
We can then combine like terms. Distributing the 11 Ohms’s and combining the 11 Ohms’s
I2 and the five Ohms I2 results this equation. If we combined the 12 Ohms’s I3 and the
eight Ohms’s I3 in the second equation we get this. We can then divide through the bottom
equation by five Ohms’s. Moving 4 I3 to the other side, we have solved for I2. We
can then substitute this into the equation above, resulting in a single equation with
a single unknown, I3. Performing the multiplication, we have 15 volts -64 Ohms I3 -11 Ohms I3.
Combining the I3 terms, we have 15 volts -75 Ohms’s times I3. Solving for I3 gives us
I3=1/5 of an amp. Which from our previous equations is also equal to I4. Remembering
that I2 is equal to four times I3, gives us that I2 is 4/5 of an amp. Remembering that
I1 is then equal to I2 + I3 gives us I1 is 5/5 of an amp. Which is also equal to IS.
We can then substitute these known values into our Ohms law relationships to solve for
the four voltages. To make this a little more practical, let’s first convert the fractions
to decimals. It is then just a matter of multiplication to solve for the four voltages. So the values
for all our currents and voltages are summarized here. The values drawn back into the schematic
would look like this.
Hopefully, at this point, we can recognize at least two things: First, using Ohms law,
KCL, and KVL, we can solve any circuit that contains sources and resistors. Second, that
even for a relatively simple circuit this is a fairly long and tedious process. The
first realization gives us the confidence that we can solve any resistive circuit. The
second realization probably makes us hope that there is a more efficient to do this.
Thankfully, there are many other ways. By applying these three laws to several different
situations, we will begin to notice and patterns and from those patterns we can develop more
efficient ways of solving circuits.
In the next couple of videos we’ll define series and parallel connections and, from
there, see if we can notice anything about those arrangements that could make the process
of solving a circuit less painful.
That’s all for today, until next time, go out and make it a great one.