### Video Transcript

The current ๐ผ in the circuit shown
is 2.5 milliamps, which is the greatest current that can be measured using the
circuit as an ammeter. The resistance of the galvanometer
is 10 times the resistance of the shunt. What is the difference between ๐ผ S
and ๐ผ G? Answer to the nearest
microampere.

Letโs start by writing down the
information weโve been given in the question. First, we were told that the
current ๐ผ equals 2.5 milliamps. We also know that the resistance of
the galvanometer, which weโll call ๐
G, is 10 times the resistance of the shunt ๐
S. So in equation form, we can say
that ๐
G equals 10๐
S, where ๐
G is the resistance of the galvanometer and ๐
S
is the resistance of the shunt.

The question is asking us to find
the difference between ๐ผ S and ๐ผ G. In other words, we want to find the
current through the shunt minus the current through the galvanometer. To do this, we should also recall
some more general equations that apply to this circuit. To start, letโs think about the
current at this node here. We can see that the current ๐ผ
splits into the two other currents, ๐ผ G and ๐ผ S, as shown by the red labels on the
diagram. Since charge is conserved, we know
that total current ๐ผ must equal the sum of these two currents, so ๐ผ equals ๐ผ G
plus ๐ผ S.

Next, because parallel branches in
a circuit have the same potential difference across them, we can say that ๐ S, the
potential difference across the shunt, is equal to ๐ G, the potential difference
across the galvanometer. To expand upon this, letโs apply
Ohmโs law, which states that potential difference equals current times
resistance. This tells us that ๐ S equals ๐ผ S
times ๐
S, and ๐ G equals ๐ผ G times ๐
G. Subbing these in, we have that ๐ผ S
times ๐
S equals ๐ผ G times ๐
G.

Now, we know a useful relationship
between these resistance values. So to minimize the number of
different variables in this expression, letโs substitute 10๐
S in for this ๐
G
term. And since weโre trying to find a
relationship between these two current values, next we should make one of them the
subject of this equation. We could choose either one. So for now, letโs just choose ๐ผ S,
the current through the shunt. Dividing both sides of the equation
by ๐
S and then canceling the like terms out of the numerator and denominator of
both sides, we have that ๐ผ S equals 10 times ๐ผ G.

Now that we have the current
through the shunt expressed in terms of the current through the galvanometer, letโs
substitute this into our expression for the current into and out of the
junction. So we have that the total current
๐ผ equals ๐ผ G plus 10๐ผ G or simply 11๐ผ G. And remember that we have an actual
numeric value for the total current ๐ผ, which is 2.5 milliamps. So letโs rearrange this to solve
for ๐ผ G in terms of ๐ผ. Dividing both sides of the equation
by 11, we know that the current through the galvanometer is one eleventh of the
total current. Since we know that ๐ผ S equals 10๐ผ
G, we can also say that ๐ผ S equals ten elevenths of the total current.

Finally, we have the current
through both the shunt and galvanometer expressed in terms of a value that we
know. So weโre ready to find their
difference, ๐ผ S minus ๐ผ G. Substituting in our expression for
each term and simplifying, the difference can be written as simply nine elevenths of
the total current ๐ผ. So, plugging in the value of the
total current, 2.5 milliamps, and grabbing a calculator, we get a result of about
2.045 milliamps. All we have to do now is express
this in microamperes, and weโve reached our final answer.

Thus, we found that the difference
between the current through the shunt and the galvanometer equals 2045
microamperes.