Hello everyone. I'm Jesse Mason and for this
edition of Teach Me we're going to use Kirchoff's Rules to analyze a circuit. Specifically,
we're going to determine the current through, the voltage across and power dissipated by
the two resistors you see here. As always, when using Kirchoff's Rules, we're going to
start by labeling our junctions. I'm going to label mine J1 and J2. Next we label our
currents. I'm going to label this current I-0. After Junction 1 I'll have I-1 on the
right leg. Note that I-0 does not pass through Junction 1. This middle leg will be I-2. And
when I-1 and I-2 rejoin we have I-0, because the current coming out of that 1.5-volt battery
needs to be the current coming back in. Also note that the directions of these currents
at this point are arbitrary; we'll find out if we chose the right directions after we
complete the problem. Next we label our loops. I'll label this Loop A and this Loop B. Just
like the currents, the direction of your loops is arbitrary. And you can label this outer
perimeter Loop C, if you'd like. I'm not going to -- I anticipate that we won't need it.
Now that we've labeled our circuit we're going to apply the Junction Rule. The Junction Rule
states that the sum of the currents into a junction is equal to the sum of the currents
out of a junction. So we're going to apply this to Junction 1 and the currents in is
just I-0. So on the left hand side of the equation I'm going to write I-0. And the currents
coming out of that junction: we have I-1 here and I-2 here. So on the right side of the
equation we'll have I-1 plus I-2. This Junction Rule, by the way, is just a consequence of
the Conservation of Charge. Now if I apply the Junction Rule to Junction 2, we'll have
I-1 coming in, as well as I-2 coming in -- so on the left side of the equation we'll have
I-1 plus I-2 -- and we have I-0 coming out. So on the right side, we have I-0. This is
exactly what we have already written so we're not going to use that equation. Now we apply
the Loop Rule to Loop A, which states that the sum of the voltages around a closed loop
is equal to zero. Starting in the upper left hand corner, we're going to move clockwise
around Loop A. And the first component we get to is the 100-ohm resistor. Now here we're
traveling clockwise so we're coming down this leg moving with the direction of the current.
This means we're gonna have a voltage drop across the resistor. Negative I R, so I'm
going to write negative I-2 times 100 ohms to indicate the voltage drop. And we continue
to move clockwise around Loop A until we get to our next component, which is the 1.5-volt
battery. Here we're moving from low to high, negative to positive. This indicates a voltage
lift. Positive V. So I'm going to write plus 1.5 volts. And that's the last component in
this loop so I set it equal to zero. Next we use the Loop Rule to analyze Loop B. Starting
again in the upper left hand corner we travel around clockwise until we get to the first
component, a 9-volt battery. Traveling from high to low, that means a voltage drop - minus
9 volts. Continuing around Loop B we get to the 200-ohm resistor. We're traveling clockwise,
which is with the direction of I-1. That's gonna indicate a voltage drop -- minus I R.
So I write minus I-1 times 200 ohms. And then we continue around Loop B, up the middle leg
until we get to the 100-ohm resistor. Here we're traveling up the leg whereas the current
is traveling down. This is gonna indicate a voltage lift -- plus I R. So I write plus
I-2 times 100 ohms. Complete Loop B, set this equal to zero. And at this point the physics
of determining the current in this problem is done. All that's left is some algebra.
We have three equations, three unknowns. We can solve this. I'm gonna start with the middle
equation here. Solve for I-2. Now I'm gonna do something a little unorthodox for physicists:
I'm going to drop the units. This is pretty much the only time I ever do this, but when
using Kirchoff's Rules, it makes the equations much easier to handle, especially when they
get really hairy. So this is what my equation looks like. I'm gonna solve this for I-2.
I-2 is equal to 0.015 amps, or 15 milliamps. I box this up. And now I'm gonna use this
value to solve for I-1 in the equation above. I'm gonna plug it in here and solve for I-1.
Again, I'm going to drop my units. Rewriting the equation. By the way, if you're comfortable
with linear algebra you may want to set these equations up in a matrix and solve for the
variables that way. Often that is a much simpler way of handling the algebra. In this case,
I'm just gonna plug and chug. So this product is equal to 1.5 volts. Move it to the other
side, I have negative 200 I-1 is equal to positive 7.5. So we have I-1 equals 0.0375
amps. Oh, don't forget this negative sign here. That's important - negative 0.0375 amps.
Or negative 37.5 milliamps. So what does this negative sign tell us? This negative sign
tells us that the direction that we chose for I-1 is the wrong direction. Remember how
we picked it arbitrarily at the beginning of the problem? Well, here's where we find
out that the actual direction of positive charge flow is the other way. No harm, no
foul. Now we know the correct direction of the current. So we're gonna take I-1 and I-2,
plug them into the Junction Rule to determine I-0. So I-0 is equal to negative 0.0375 plus
0.015. So we have I-0 is equal to negative 0.0225 amps. Or negative 22.5 milliamps. I'm
gonna box this up. And again what this negative sign is telling us is the direction that we
assigned for current is the incorrect direction of the positive charge flow. It's actually
moving the other way. So now that we have the currents in this circuit we can determine
the voltage drops across the resistors using the current and the resistance values. In
other words, we're gonna use Ohm's Law. So for current we use the 15-milliamp value and
the 100-ohm value. And we find out that the voltage drop across the 100-ohm resistor is
going to be 1.5 volts. I'm going to do the same thing for the 200-ohm resistor but to
conserve some space here I'm gonna skip the calculation and just skip to the value. So
using that current at the 200-ohm resistor, we have a value of 7.5 volts for the voltage
drop across the 200-ohm resistor. To determine the power dissipated by these resistors, we
simply multiply the current through by the voltage across -- I V. And we find out, for
the 100-ohm resistor, we use the 15 milliamps for the current and 1.5 volts that we just
determined. And we find out that the power dissipated by the 100-ohm resistor is going
to be 0.225 watts, 22.5 milliwatts. And to determine the power dissipated by the 200-ohm
resistor we're just gonna multiply the current times its *voltage* value, which yields 281.25
milliwatts. And I'll box this up. And we're done. So there you have it: the current through,
the voltage across and the power dissipated by resistors using Kirchoff's Rules. I'm Jesse
Mason. I hope this was helpful to you. And until next time, happy learning!
