in this video we're asked to solve a resistor network by using kirchhoff's laws
and the reason we're using kirchhoff's laws here is because the resistors are not in series or
parallel to make it clear why this is true let's look at these two resistors if you think they're
in parallel well we don't have a problem with this side of them that's connected directly by wire
but if i look at the other side there's an extra power source in the way here
so the voltage can't be the same across each of these resistors
which is the defining characteristic of parallel resistors so they're not parallel if you think
they're in series that's wrong because whatever current is passing through this one gets split
at this node so the two resistors can't possibly have the same current going through them which is
the defining characteristic of series resistors so we're stuck having to use a different tool
set and that's kirchhoff's laws so what i'm going to do here is start with kirchhoff's voltage law
and i'm going to make sure that i touch every single circuit element
with my voltage loops because i know my equations must involve every resistor and power source
so i'll just go ahead and call this loop one we're going to stay inside that upper loop
i'll call this loop two the lower left loop and loop three the lower right loop in each of these
loops i'm going to start at the lower left hand corner to keep myself organized and then i have
to remember the rules for voltage increases and decreases if i'm going the same direction as the
current across a resistor that must be a voltage drop if i'm going the opposite direction it would
be a voltage increase if i go from the negative to positive terminal of the battery that's a voltage
increase and vice versa and now i can write down my first three equations as i travel through my
first loop i go the opposite direction of current over this 4 ohm resistor so i'm going to have a
voltage increase there the magnitude is given by ohm's law that's v equals ir so i have plus
4 i4 then i cross over this 2 ohm resistor in the opposite direction of i2 so i get plus 2 i2
and then i cross over this 5 ohm resistor in the same direction as i3 so that's negative 5i3
and i'm back where i started so the sum of all those changes must be zero that's just saying
if you make a complete loop you end up back at the same potential you were to begin with
number two i cross over this 2 ohm resistor same direction as i six i cross over this 2 ohm
resistor opposite direction of i5 that's a voltage increase to i5 i cross over this 4 ohm resistor
same direction as i 4 and finally i go from the high side to the low side of this
power source so that's -2 volts in loop 3 the first thing that i hit
i go from the low side to the high side of this power source that's 6 volts
the second thing i hit same direction as i2 across the 2 ohm resistor
and the final piece here same direction as i5 across the 2 ohm resistor
and i'm back where i started next i use kirchhoff's current law which says that all the
current entering a node must add up to 0. another way of saying that is that all the currents into a
node must be equal to all the currents out of the node so i'm going to label these nodes number four
and five and six and look at all the currents coming in and out that'll
give us six equations for our six unknowns and then we'll go to the computer algebra system
at node four i4 is coming in and i3 is coming in i6 is coming out at node five i have i2
coming in i4 and i5 coming out at node six i have i1 coming in i2 and i3 coming out
all right so now we're going to plug all these equations into a computer algebra system
and solve the system all right we're working in a computer algebra system called wx maxima
here this is an open source computer algebra system and i'll post a link to the download page
in the description i've entered all my equations with the right syntax and i hit shift enter
and maximum is going to repeat these equations to me and i can just double check that i've done
them all right everything looks good to me so now i'm going to solve the system so there's
the syntax you solve your set of equations for your set of variables hit shift enter
and notice that maxima spits out exact fractional solutions for these i'm not really interested in
that i'd rather have decimal approximations so i'm going to ask maxima to round the prior result
and now i've got my decimal approximations
rounding all these to three sig figs so we've got every branch current in the circuit note that one
of these is negative i4 came out negative and all that means is that the actual direction of i4 is
opposite to the way it was drawn finally we're asked to get the potential difference from node a
to node b so all i need for this is the voltage drop across this 2 ohm resistor now that i have
the current through that resistor it's no big deal v a b is going to be the voltage
drop across that resistor it's given by ohm's law i times r the current is 1.51 amps
the resistance was 2 ohms and i get 3.02 volts so in the lab i can test this by just putting
the multimeter probes at these two junctions in the circuit and it should get very close to
3 volts if you find the physics content on zack's lab helpful click on the zaxlab logo
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