in this video we introduced the idea of a non-ideal power source and the terminal voltage
of the non-ideal power source so until now we've treated batteries as if they have no internal
resistance in fact all batteries have some amount of internal resistance and we can model any
battery as an ideal power source sometimes called an emf or electromotive force in series with a
small internal resistance now the nodes that we've labeled here in the circuit a and b those are the
actual battery terminals if the battery isn't connected to anything then no current can flow
and if we put our multimeter across these nodes a and b we're just going to read exactly the same
voltage as the emf that's being maintained by the chemical reactions in the battery however
when we connect this battery to a resistance and i'm calling that the load resistor in this case
then i'm going to have a current flowing through the circuit and that means there's a current
flowing over the internal resistance of the battery well ohm's law tells us that if a current
is flowing over a resistor we have a voltage drop and this means the voltage between a and b is
going to be less than the ideal emf so the way we sort of formalize these voltage differences
is i would take a path from one node to the other and i'll say the potential at a plus
the emf as i'm traversing this path up the left branch of the circuit i just went up by a voltage
equal to the amf minus ir so i applied ohm's law there i said v equals ir for the resistor and that
was a voltage drop so va plus the emf minus ir is equal to vb this allows us to express
vb minus va as the emf minus ir and this difference vb minus va that's the voltage
across the terminals of the battery and we call that the terminal voltage sometimes we'll use a vt
for that so we finished with a typical example of how you might be asked about the terminal voltage
we're told that a battery produces an emf of 9.0 volts and has an internal resistance of 0.1 ohms
my software automatically turned that into 100 milliohms but we'll just change it back
and we want to compute the terminal voltage of the battery when it's connected to a 47 ohm
load resistor so all we're going to do here is simplify the circuit to have a single resistor
and then we're going to find the total current and then that will allow us to find the voltage drop
over the internal resistance and we can get our terminal voltage for the battery so we'll redraw
the circuit with a single resistor all i have is two resistors in series so it's just an r1 plus
r2 kind of thing and again 100 milliohms is just 0.1 ohms so when i add these i get 47.1 ohms now
i apply ohm's law and i say that i is equal to v over r for this that's 9 volts over 47.1 ohms
and i find that 0.191 amps are flowing through the circuit if i then go back to my original circuit
i have i equals 0.191
amps and then i can get the voltage drop over the internal resistance
and get my terminal voltage so my terminal voltage is going to be the ideal voltage of 9 volts
minus the voltage drop due to the current over the internal resistance 0.191 amps
times 0.1 ohms sometimes that's called an ir drop because we just took current times resistance to
get it and i get 8.98 volts out of this so again anytime you have a battery that's actually running
a current through a circuit the terminal voltage on that battery is going to be less than what
it is if it's not connected and the bigger the current that you're running through the battery
the bigger the drop over that internal resistance so the terminal voltage is going to drop
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