in this video we're given a resistor network circuit and this circuit has two power sources
in it and it has a bunch of resistors that are not in series or parallel so just as a quick reminder
of how i know these resistors are not in series or parallel if i look at maybe this node right
here and this node right here there's a 5 ohm resistor and a 2 ohm resistor between those nodes
but clearly the current doesn't have to be the same through each of those resistors so they can't
be in series but i can also tell the potential difference doesn't have to be the same across
each of these resistors because they're not directly connected by wire at each side of each
of the resistors there's an extra power source in the way in this middle branch so those two
resistors are not in series or parallel there's many other examples throughout the circuit so
we have to take a new approach to the problem and that approach is kirchhoff's laws and there's two
different basic ideas with kirkhof's laws we can analyze loops within a circuit and it's a really
common sense idea if i go through a closed loop in a circuit from a starting point all the way back
to that starting point then the sum of all the voltage changes on that closed loop must be zero
the second part of kirchhoff's laws is to look at the nodes in a circuit and that's all the
points where branches are connecting within the circuit that's these white dots within the circuit
and again it's a really common sense idea what kirchhoff's laws say here all the current
entering a node must be equal to the amount of current exiting a node so you can't have charge
just building up at some point in a circuit the amount going in has to be equal to the amount
going out so when we analyze the circuit using kirchhoff's laws we label all the branch currents
so every branch in the circuit gets labeled with a current i've already done this in this problem
i have i1 through i6 for the six distinct branches in the circuit so anywhere where current is
allowed to split or combine you have to label a new branch current and then the loop analysis
is a way of generating equations that relate these unknown currents and the node analysis
can give you more equations and ultimately what we're looking for is six independent equations for
these six unknowns now when i teach physics i'm definitely not interested in seeing if students
can solve a system of six equations six unknowns by hand and that's why we're instructed to use a
computer algebra system to get the work done here so let's begin with our loop analysis
the most obvious choice of loops here is to make this upper left rectangle a loop make the
upper right rectangle a loop and make the lower rectangle a loop and i indicated that i'm going
to traverse these loops in a clockwise direction there are additional loops that you can construct
for example the entire outer edge of this circuit would give you another loop or i could do a loop
where i was just using the first half of the upper part cutting down to the middle branch
and then going all the way around the bottom so you have a lot of flexibility in your choices here
but my intent in this problem is to illustrate both parts of kirkhof's laws basically with equal
weight so i'm going to choose three loop equations and then i'll generate three node equations
so we analyze voltage changes around loop one and that requires picking a starting point
i'm going to start at the lower left corner and again we're going clockwise around this thing
the first thing i notice is that i cross a 3 ohm resistor in the same direction as the branch
current i4 if i'm going across the resistor in the same direction as i-4 that means i have a voltage
drop and that's given by ohm's law v equals ir so i just take the resistance and multiply by
the current and it gives me a drop of negative 3 i4 note that if we've drawn i4 in the wrong
direction the answer will simply come out as a negative number at the end so you don't have
to worry about guessing the directions correctly next i'm going to go through this vertical branch
where i3 passes through a 2 ohm resistance that's the same direction as i3 so that's a voltage drop
and that will give me a negative 2 i3 finally i start to go along the bottom branch of this
loop and i have a current of i6 passing through a resistance of 1.5 ohms again that's a voltage drop
because i'm going the same direction as the current the way it was drawn
and then the last circuit element that i go across is this battery and i'm going from the low side to
the high side so that's an increase of 10 volts those are all my voltage changes and by the time
i get back to where i started the sum of all the changes must be zero and i have my first equation
next we move to loop two and i'll go ahead and start at the lower left corner of this one as well
and we're moving through this in a clockwise direction and the first thing i do is cross this
two ohm resistor in the opposite direction of i3 that means i must be going up in voltage so this
one's going to give me a positive 2 i3 applying ohm's law then i get to this 5 ohm resistor
and i can see that the branch current i2 is just going this way by the time it loops around to the
top of the circuit that i2 must be the same as this one because there's no way it could split
between this part of the circuit and this part of the circuit and when i cross that 5 ohm resistor
i'm going the opposite direction of i2 so again it's a voltage increase and that's a plus five
i2 then i continue to travel clockwise around this loop and the last circuit element that i hit
is this six volt power source and i'm going from the high side to the low side which means i have a
negative change in voltage so that's a minus six and i have my second loop equation my third
and final loop i'll start that at the lower left corner just to be consistent and so i'm traveling
this way and the first thing i hit is this 10 volt power source and i'm going from the high side to
the low side so that's a voltage decrease of 10 volts then i hit the 1.5 ohm resistor opposite to
the direction of i6 so that's a voltage increase of plus point five i six next i'm going from the
low side to the high side of that six volt power source so that's a plus six volts and then finally
i'm going through the six ohm resistor opposite to the direction of i1 so that's a voltage increase
of plus 6i1 and i have my final loop equation now let's get into the nodes so we have four nodes in
this circuit to choose from but we only need to generate three equations so we're just going to
pick three of these to generate our three additional equations and then we'll be done
just a word of warning about these node equations if i were to generate
all four of these node equations i'm guaranteed to get one of those equations that's dependent
on the other three and basically the idea is if i count up all the ways the current can split apart
and all the ways the current can come together i'm going to end up double counting somewhere
so only three of these are going to be useful anyway
and i'll go ahead and just pick the three nodes that lie along the central branch
so my first one will be this one and i'll call that node four and then i just look for what's
flowing into node four and what's flowing out of node four so i see i6 is flowing into node 4
but i1 and i4 are flowing out so i6 goes in i1 and i4 go out and that gives me a simple equation
i6 equals i1 plus i4 i'll call the node in the middle node 5 and what i see flowing into that
is i3 and flowing out of that i6 and i5 finally i'll label node 6 in the middle on the right
and what i see flowing into that is i5 and i1 so i1 just turns around that corner at the bottom
right side of the circuit it flows into node 6. so i5 and i1 are flowing in i2 is flowing out
okay so now we're set up with a system of six equations and six unknowns
this is very much an appropriate application for a computer algebra system and the computer algebra
system we're using today is called wx maxima this is a free open source computer algebra system
and i'll post a link to the download page in the description what i've done here already
is define my equations so i just gave names to each of these equations just to simplify my life
so equation one two three four five and six those should all look the same as the equations
we wrote down by hand so we can just double check real quick that everything is correct
and then we get into actually solving the system of equations so i'm going to hit shift enter and
that defines these names for each of the equations and now we can get into solving the system so what
i want in the end is decimal approximations for all six of the branch currents so i'm going to
start by using the word float which tells maxima to give us a floating decimal point
answer instead of an algebraic answer then we're going to solve the system with the solve command
and we just put in brackets the name of the equations we want to solve and then we put
in brackets the names of the variables we want to solve for and we're going to hit shift enter
and when i do this the decimal approximations are all going to pop out and i'll go ahead and write
these all down to the thousandths place i1 is 0.267 amps i2 is 0.477 amps three is 1.807 amps
and so on now i want to make one final point before i wrap things up measuring currents is
very difficult to do in the lab what we would ordinarily do if we had this circuit set up on
a breadboard is we would be using a multimeter to measure voltage differences between nodes so
knowing these branch currents allows you to find voltage differences between any two nodes in the
circuit for example if i looked at this one at the top middle and then this one in the very middle
of the circuit if i wanted to predict a potential difference between those points
all i need to use is my numerical result for i3 i would then use v equals ir to find the voltage
drop over that 2 ohm resistor that would give me a prediction for the node voltage between those
two nodes and then i could test that in the lab if you enjoyed this video or at least found it useful
check out another one by clicking one of the links on the left or click the zaxlab logo on
the right to explore dozens of physics and math playlists as always you can leave your questions
comments and requests in the comments section below and i'll get back to you within 24 hours
thanks for watching zach's lab and best of luck on your math and physics journey