# Question Video: Calculating the Value of a Multiplier Resistor Physics

A voltmeter can measure a maximum potential difference of 4 V and has a resistance of 3,000 ฮฉ. When a multiplier resistor ๐_(m) is connected to the voltmeter in series, its measuring range increases by 12 V. Calculate the value of ๐_(m).

03:03

### 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 ๐ผ sub G is the maximum current that the galvanometer scale can measure and ๐ sub G is the galvanometerโs resistance, then according to Ohmโs law, this product of ๐ผ 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 ๐ผ 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, letโs 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 ๐ผ sub G equals four volts times one divided by ๐ sub G. Our galvanometerโs resistance is given to us in the problem statement as 3,000 ohms. ๐ผ 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 ๐ผ 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 ๐ผ sub G. Itโs 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 weโre effectively multiplying by one, we arenโt 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.