### Video Transcript

Which of the following is the most
correct description of how the range of values of currents that a galvanometer can
produce is extended when it is converted into an ammeter using a shunt resistor? (A) A shunt resistor with a
resistance much smaller than that of the galvanometer is connected in parallel with
the galvanometer. (B) A shunt resistor with a
resistance much greater than that of the galvanometer is connected in parallel with
the galvanometer. (C) A shunt resistor with a
resistance much smaller than that of the galvanometer is connected in series with
the galvanometer. (D) A shunt resistor with a
resistance much greater than that of the galvanometer is connected in series with
the galvanometer. (E) A shunt resistor with a
resistance equal to that of the galvanometer is connected in parallel with the
galvanometer.

In this question, we want to work
out where we can place the shunt resistor in the circuit in order to extend the
range of currents that the galvanometer can produce. Letβs begin by connecting a shunt
resistor in series with the galvanometer, as is described in options (C) and
(D).

Recall that when we have two
resistors connected in series, the total resistance is given by the sum of the
individual resistances. So, by labeling the resistance of
the galvanometer as π
sub πΊ and the resistance of the shunt resistor as π
sub π,
we have that the total resistance π
sub total is equal to π
sub πΊ plus π
sub
π.

Also, recall that Ohmβs law can be
written as π equals πΌ times π
, where π is the potential difference, πΌ is the
current in the circuit, and π
is the resistance. We can rearrange this to get an
expression for the current in the circuit. To do this, we just need to divide
both sides of the equation by the resistance π
. This gives us the expression πΌ is
equal to π over π
. Applying Ohmβs law across this
circuit, we get πΌ is equal to π over π
sub total.

If we pick a shunt resistor with a
resistance much smaller than that of the galvanometer, that is, when we pick a shunt
resistor such that π
sub π is much less than π
sub πΊ, then π
sub total is equal
to π
sub πΊ. In this case, the shunt resistor
has pretty much no effect on the resistance of the circuit, meaning it does not
change the range of current values that the galvanometer can produce. So, option (C) is incorrect.

If we pick a shunt resistor with a
resistance much larger than that of the galvanometer, that is, if we pick a shunt
resistor such that π
sub π is much greater than π
sub πΊ, then π
sub total is
equal to π
sub π. In this case, the denominator in
our expression for the current will be much larger than before. This means that adding this
resistor in series will make the current much smaller. In other words, this decreases the
range of current values that the galvanometer can produce. So, option (D) is incorrect.

So now, letβs try connecting the
shunt resistor in parallel with the galvanometer, as is described in options (A),
(B), and (E). Recall that the current splits
across parallel paths. We can apply Ohmβs law to each path
separately to find out how much current is in each path. Weβll call the current through the
galvanometer πΌ sub πΊ and the current through the shunt resistor πΌ sub π. So, the currents through the
galvanometer and shunt resistor are πΌ sub πΊ is equal to π sub πΊ over π
sub πΊ
and πΌ sub π is equal to π sub π over π
sub π, respectively.

We know that both paths have the
same potential difference, so this means π sub πΊ equals π sub π, which is equal
to π, the potential difference provided by the cells. So, the current through the
galvanometer is equal to π over π
sub πΊ, and the current through the shunt
resistor is equal to π over π
sub π.

Letβs consider what happens when we
choose a shunt resistor with a resistance much smaller than that of the
galvanometer, that is, when we pick a shunt resistor such that π
sub π is much
less than π
sub πΊ. Since πΌ sub πΊ and πΌ sub π have
the same values in the numerator, this means that if π
sub π is much less than π
sub πΊ, then πΌ sub π is much greater than πΌ sub πΊ. In other words, most of the current
goes through the path containing the shunt resistor.

Meanwhile, there is a small current
through the galvanometer. The current that passes through the
galvanometer is a constant proportion of the current in the circuit. This means that the deflection of
the galvanometerβs needle will be proportional to the current in the circuit. So, if a large current is passed
through the circuit, only a small proportion of it will pass through the
galvanometer. This means that a much larger
current can pass through the circuit before the galvanometer reaches full-scale
deflection.

Hence, the combination of the
galvanometer and shunt resistor in parallel may be used to extend the range of
values of currents that a galvanometer can produce. Therefore, option (A) is the most
correct description. A shunt resistor with a resistance
much smaller than that of the galvanometer is connected in parallel with the
galvanometer.