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!