### Video Transcript

A galvanometer has a coil of
resistance 25 ohms. What is the value of the resistance
of the multiplier resistor that is required to allow the galvanometer to measure
potential difference that is five times the potential difference across its
coil? (A) Five ohms, (B) 20 ohms, (C) 100
ohms, (D) 125 ohms.

The question wants us to calculate
the value of the multiplier resistance combined with a galvanometer used to convert
it into a voltmeter. Letβs recall that a multiplier
resistor must be connected in series with the galvanometer in order to convert the
galvanometer into a voltmeter. The question tells us the
resistance of the galvanometer. Labeling this as π
sub G, we have
that π
sub G is equal to 25 ohms.

We are also told the maximum value
of the potential difference that the voltmeter to be constructed must be able to
measure. This maximum measurable potential
difference, which weβll call π sub max, is five times the potential difference
across the galvanometer coil, which weβll call π sub G.

We want to find the value of the
multiplier resistance π
sub M. To do this, we can start by
recalling that thereβs an equation for the maximum value of potential difference the
voltmeter can measure, which comes from Ohmβs law. This equation says that π sub max
is equal to the current πΌ sub G through the galvanometer multiplied by π
sub M
plus π
sub G. On the left-hand side of the
equation, we know that π sub max is equal to five times π sub G, where π sub G is
the potential difference across just the galvanometer coil.

Using Ohmβs law again, π sub G can
be written as π sub G equals πΌ sub G multiplied by π
sub G. So then π sub max, which is equal
to five times π sub G, must be equal to five times πΌ sub G times π
sub G. We then have that π sub max is
both equal to five times πΌ sub G times π
sub G and to πΌ sub G multiplied by π
sub M plus π
sub G. Therefore, it must be true that
five times πΌ sub G times π
sub G is equal to πΌ sub G times π
sub M plus π
sub
G.

Expanding out the bracket on the
right and then subtracting πΌ sub G times π
sub G from both sides, we have that
four times πΌ sub G times π
sub G is equal to πΌ sub G times π
sub M. From here, we see that the factor
πΌ sub G appears on both sides. So we can cancel this out. We have then that π
sub M is equal
to four times π
sub G. We know that π
sub G is equal to
25 ohms.

Substituting this into the equation
for π
sub M, we have that π
sub M is equal to four times 25 ohms. This works out as 100 ohms. We can see that this value matches
the one given in option (C). The correct answer is therefore
option (C). The resistance of the multiplier
resistor must be 100 ohms.