in this video we introduce a simple device called the galvanometer and we explore how the
galvanometer can be used to construct an ammeter or a voltmeter a galvanometer consists of a needle
connected to an axle with coils of wire that rotate with the needle in addition there's a
permanent magnet exerting a magnetic field across the coils and in this picture that magnetic field
would point left to right the needle points at a scale which right now is pointing at
zero and the spring will return the needle back to zero after it's deflected so when the galvanometer
is plugged into a potential difference current will flow through the coils creating a magnetic
field in the coils those coils then interact with the magnetic field from the permanent magnet
and the larger the current the larger the torque exerted on the needle and it will deflect further
along the scale so in our two examples we're going to look at how do we use a galvanometer to make
a volt meter and how do we use a galvanometer to make an ammeter and in each case the galvanometer
has the same specs so i'm just going to list them over here next to our galvanometer picture
first we're told that the needle reaches full deflection when .006 amps flows through the coils
of the galvanometer in other words 0.6 milliamps second we're told the galvanometer comes with
internal resistance that's the resistance of those coils of wire of 0.4 ohms so let's take a
look at how we could use this as a voltmeter what we're told is that we want to design a voltmeter
that's fully deflected when it's connected across a voltage of 10 volts in other words when we hook
it in parallel across this potential difference of 10 volts we get full deflection that is a current
of .006 amps may as well label the internal resistance here that was 0.4 ohms
and so i can imagine if i just hook the galvanometer up to this 10 volt potential
difference without adding some extra resistance to it i would end up pulling a huge current i would
end up with i equals v over r just using ohm's law 10 volts over 0.4 ohms internal resistance
and we'd end up pulling 25 amps of current through our galvanometer and the thing would burn up
so what we're going to do to prevent this and to control the maximum current is put a resistor in
series with the galvanometer such that when we connect to a 10 volt potential difference we end
up with 0.0006 amps of current running through it okay so that 25 amps is a bad thing and we're
changing our strategy here i'm going to apply ohm's law to the galvanometer branch v equals ir
and i know it has a 10 volt potential difference across it and i'll plug in the numbers later
and i know i is .0006 amps when it's at maximum deflection and my total resistance is big r that's
the resistor i'm going to put in series and little r the internal resistance of the galvanometer
distributing the big i i'm going to go ahead and solve everything symbolically here and subtracting
the i little r from both sides i get that my big r required is v minus i little r
divided by i i get 10 minus point zero zero zero six times point four ohm's internal resistance
over point zero zero zero six and i end up with sixteen thousand six hundred sixty six ohms which
i'm going to write as 16.7 kilo ohms again what we've accomplished here is we made the resistance
in the galvanometer branch just right so that if we hook it up in parallel over a 10 volt circuit
element we're going to pull a current of .0006 amps and get a maximum deflection of the needle
and then we can say when the needle is here we're looking at
10 volts deflection and this is how we can begin to calibrate this galvanometer to be a voltmeter
one final comment on voltmeter design is we want it to interfere very little with
the actual circuit that we're trying to study and so voltmeters have enormous resistances
we have 16.7 kiloohms in there which means we're barely drawing any current right it's .0006
amps and so we're affecting the circuit that we're trying to study as little as possible when
we connect this thing to it in the second example we want to use the same galvanometer as an ammeter
so ammeters need to be connected in series with a circuit so you can watch all the current flow
through the ammeter and measure how much there is and again just using this generic box i'm saying
we have two amps flowing through this circuit and i want my galvanometer to be fully deflected in
this case well if i simply put the galvanometer in series with the circuit there'd be two amps
of current going through it minus a little bit because of this internal resistance
and the thing would burn up right it reaches maximum deflection at a fraction of a milliamp
so what i do is i put a resistor in parallel with the galvanometer branch
and that resistance should be very small so that the vast majority of current goes that way
and then i end up with .0006 amps going through the galvanometer branch when there's two amps
flowing through the ammeter so my total current is 2 amps but just to stay symbolic for a second
i'm going to call this i total coming into the parallel combination and that splits into
i1 and i2 and i know what i1 is that's .0006 amps and i2 that would be 2 amps minus 0.0006
amps and i'll write it down as an equation i total equals i1 plus i2 and then i can say because this
is a parallel combination the voltage drop over the galvanometer branch must be the same as the
voltage drop over this new resistor that we're putting in so i set these two things equal and
i'm going to use ohm's law to quantify the voltage drop that's just v equals ir so i have i 1 little
r and that's got to be equal to the voltage drop across big r which is going to be i2
big r so i directly know i total in i1 so what i'm going to do is in the first equation i'm
going to solve for i2 and then sub it into the second equation and then we'll solve for big r
when i plug the numbers in so i1 that was the name for the current in the galvanometer branch
which is .0006 amps internal resistance there was 0.4 ohms and divided by i total which was 2 amps
minus i1 when i run the numbers on this i get 1.20 times 10 to the negative 4
ohms which i could write as 0.12 milliohms so just a final comment on our ammeter design
ammeters always come with a very tiny resistance in parallel with the galvanometer branch and
this makes it so you have a minimum amount of interference with your circuit i could say that
because r is a very small thing 0.12 milliohms the potential difference across our ammeter is
guaranteed to be really small and again our goal is to study the circuit without modifying it
too much so this is a good thing if you find the physics content on zack's lab helpful click on the
zaxlab logo on the right to browse playlists and subscribe to the channel i produce over 100 new
videos per month and subscribing is the easiest way to find new content thanks for watching you