### Video Transcript

A voltmeter can measure a maximum
potential difference of four volts and has a resistance of 3,000 ohms. When a multiplier resistor ๐
sub
m is connected to the voltmeter in series, its measuring range increases by 12
volts. Calculate the value of ๐
sub
m. (A) 9,000 ohms, (B) 12,000 ohms,
(C) 15,000 ohms, (D) 6,000 ohms.

In this example, we have a
voltmeter, which is actually made from a galvanometer, a device for measuring
current. If ๐ผ sub G is the maximum current
that the galvanometer scale can measure and ๐
sub G is the galvanometerโs
resistance, then according to Ohmโs law, this product of ๐ผ sub G and ๐
sub G is
equal to the voltage in the circuit, in this case four volts. To this circuit we imagine adding a
multiplier resistor in series with the galvanometer. With this component added, the
maximum current measurable in the circuit is still ๐ผ sub G. But now the total resistance in the
circuit is ๐
sub G plus ๐
sub m. Our problem statement tells us that
by adding our multiplier resistor, we increase the measurement range of our circuit
by 12 volts. If we add 12 volts to four volts,
we get 16 volts.

Knowing all this, we want to solve
for the value of the multiplier resistor. To begin doing this, letโs solve
for the maximum current our galvanometer can measure. If we multiply both sides of this
equation by one over ๐
sub G, then ๐
sub G multiplied by one over ๐
sub G
equals one. And then reversing the values on
either side of this equality, we find that ๐ผ sub G equals four volts times one
divided by ๐
sub G. Our galvanometerโs resistance is
given to us in the problem statement as 3,000 ohms. ๐ผ sub G then equals four volts
divided by 3,000 ohms.

Knowing this and moving on to our
second equation here, we can divide both sides of this equation by ๐ผ sub G so that
it cancels out on the right. And then with the expression that
remains, we can subtract the resistance of the galvanometer ๐
sub G from both
sides. On the right, positive ๐
sub G
minus ๐
sub G adds up to zero. Finally, if we swap the sides of
our equation, we have an expression where ๐
sub m, the resistance of our
multiplier resistor, is the subject. When it comes to the variables on
the right-hand side of this expression, we know the value for ๐ผ sub G. Itโs four volts divided by 3,000
ohms. And ๐
sub G itself, the
resistance of the galvanometer, is given as 3,000 ohms. Now, if we divide 16 volts by four
volts, we see first of all that the units of volts will cancel out. And 16 divided by four is equal to
four.

As a next step, we can multiply
this first term in our equation by 3,000 ohms divided by 3,000 ohms. Because weโre effectively
multiplying by one, we arenโt changing this value. Note that in the denominator, 3,000
ohms divided by 3,000 ohms is equal to one. And in the numerator, 3,000 ohms
times four equals 12,000 ohms. ๐
sub m then equals 12,000 ohms
minus 3,000 ohms, or 9,000 ohms. This is the value identified in
answer option (A). The resistance of the multiplier
resistor in this scenario is 9,000 ohms.