in this problem we're given a resistor network circuit and this is hooked up to
a battery and our battery actually has internal resistance in it that's given us 50 milliohms
and i'm going to go ahead and express that in ohms real quick that's 0.05 ohms and the terminals of
the battery are labeled as a and b and actually in part c of the problem we're going to be asked for
that terminal voltage between the two terminals of the battery that includes the voltage drop
over the internal resistance now the rest of the circuit is just four resistors and it's a
combination of series and parallel so i have three that are clearly in parallel r2 r3 and r4 and that
parallel combination is in series with r1 in part b we're asked for the total current in the circuit
in part c we're asked for the terminal voltage of the battery and then in part d we're asked
a power question we want the power dissipated by r3 so the approach to all these resistor network
circuits is to start simplifying by using the parallel and series resistor combination rules and
to each step we're going to combine some resistors using one or more of these rules and then we
redraw the circuit we get it simpler and simpler until there's only one equivalent resistance left
that allows us to compute the total current and then we work backwards through the circuit so i'm
going to start with this parallel combination r2 r3 and r4 those are all in parallel and i
remember that one over the equivalent resistance for a parallel combination is given by the sum
of the reciprocals of all these resistances and there's not room to show every little calculation
here but i plugged into my calculator 1 over r2 that's 4.7 plus 1 over 1.5 plus 1 over 1.5
i hit enter and then i took the reciprocal of the result to get the equivalent resistance
and this comes out to 0.6468 ohms just keeping a little extra precision in the middle of this
process now we redraw the circuit with the parallel combination simplified into a single
equivalent resistor and that's what it looks like now i just have three resistors in the circuit
there's the internal resistance of the battery the original 4.7 ohm resistor and then the parallel
equivalent of those three resistors that were in parallel so for the next step i recognize
that these three resistors are in series meaning that they share the same exact current and to
combine resistors in series i simply add them up so r total is going to be the sum of these three
so i'll write that as little r for the internal resistance r1 for the 4.7 ohm resistor and req
for the parallel combination and when i add all these up i get a total of 5.397
ohms now we have the total equivalent resistance for the entire circuit and i can redraw it as a
simple circuit and our ideal voltage or emf is still 1.5 volts the total resistance including
the internal resistance of the battery is 5.397 ohms and this is the point where i can start to
make progress and work my way back so having a single ideal voltage source and a single
resistor allows me to use ohm's law to compute the current and i'll just move down to part b
on this because that's where it was asked and my total current is going to be v over r that's just
turning ohm's law v equals ir around so i have a 1.5 volt emf and then i'm dividing by 5.397
ohms and this gives me 0.278
amps so that's the total current flowing in this circuit
i can visualize it like this it goes from the high side of the battery to the low side
there's my total current i now if i work my way backwards from my simplest circuit to the one
in the middle the simplest circuit resulted from the simplification of three resistors in series
and resistors in series all have exactly the same current going through them and that means that
same current i or .278 amps must be flowing through that circuit in the middle picture
this allows me to quickly answer part c i want the terminal voltage of the battery
and that's just the ideal voltage or emf of 1.5 volts minus the voltage drop that you get
over the internal resistance and that voltage drop is just given by i times r so 0.278 amps
is flowing through that internal resistance and the value of the resistance was 0.05 ohms
and i should be able to find the voltage drop over that internal resistance
when i run the numbers on this i get .0139 volts so the terminal voltage is just going to be my
ideal voltage minus that internal voltage drop if i want to get more formal about it
this is how you could reason through it i would start at a and i would say va plus the rise in
voltage across the emf that's 1.5 minus the voltage drop across the internal resistance
has to be equal to vb and the terminal voltage is formally
vb minus va and we normally just call that vab
and it should agree with your intuition that's just the ideal voltage minus the drop over the
internal resistance 1.5 minus 0.0139 and when i run the numbers on this to three significant
digits i get 1.49 volts so there's just this little bit of a drop because of the internal
resistance finally in part d we want the power dissipated by r3 and there's more than one way
to go about this but the most efficient way to do it is to just find the voltage drop across these
parallel resistors so in my middle picture i have this equivalent resistance of 0.6468
ohms that was the three parallel resistors put together what those share in common is having
exactly the same voltage drop across them so all i have to do is get the voltage drop across this
equivalent resistor and that is going to be equal to the voltage drop across each of these in the
parallel combination in particular it will be the same as the voltage drop across r3 which is the
one i'm interested in now once i have r and v for that resistor i can get the power so how do i get
the voltage drop across the parallel equivalent resistance i have the current as 2.78 amps
in this intermediate circuit the current cannot be split anywhere it all flows
through that equivalent resistor and i have the resistance of that equivalent resistor 0.6468 ohms
into three sig figs i get .180 volts for the potential difference across those parallel
again that's the same as it is for each individual one of those parallel resistors so now that i
have the resistance of r3 that's 1.5 ohms and the voltage drop across it that's 0.180 volts
i can compute the power most fundamentally the way we compute power is by taking i times v
for whatever circuit element we're studying but for resistors in particular you can use
ohm's law to substitute into this so the thing i don't have for this resistor is i i could quickly
compute it of course by using ohm's law back at this step but to avoid that i can sub out
and replace i with v over r using ohm's law when i do that i get this power formula v squared over r
and that formula applies only to resistors i plug in my potential difference .180 volts that's going
to be squared divide by the resistance 1.5 ohms and when i run the numbers on this i get .0216
watts now it's common when we're studying circuits to move our answers into a little
nicer form by using si prefixes so it would be more common to see this answer as 21.6
milliwatts and we're done if you enjoyed this video or at least found it useful
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