Question Video: Determining the Value of the Resistance of the Multiplier Resistor | Nagwa Question Video: Determining the Value of the Resistance of the Multiplier Resistor | Nagwa

# Question Video: Determining the Value of the Resistance of the Multiplier Resistor Physics • Third Year of Secondary School

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A galvanometer has a coil of resistance 25 Ξ©. 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?

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### 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. Letβs 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 weβll call π sub max, is five times the potential difference across the galvanometer coil, which weβll 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 thereβs an equation for the maximum value of potential difference the voltmeter can measure, which comes from Ohmβs law. This equation says that π sub max is equal to the current πΌ 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 Ohmβs law again, π sub G can be written as π sub G equals πΌ 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 πΌ sub G times π sub G. We then have that π sub max is both equal to five times πΌ sub G times π sub G and to πΌ sub G multiplied by π sub M plus π sub G. Therefore, it must be true that five times πΌ sub G times π sub G is equal to πΌ sub G times π sub M plus π sub G.

Expanding out the bracket on the right and then subtracting πΌ sub G times π sub G from both sides, we have that four times πΌ sub G times π sub G is equal to πΌ sub G times π sub M. From here, we see that the factor πΌ 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.

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