(II) A milliammeter reads 25 mA full scale. It consists of a 0...

Question

(II) A milliammeter reads 25 mA full scale. It consists of a 0.20-Ω resistor in parallel with a 33-Ω galvanometer. How can you change this ammeter to a voltmeter giving a full scale reading of 25 V without taking the ammeter apart? What will be the sensitivity (Ω/V) of your voltmeter?

Accepted Answer

Welcome back. Everyone in this problem. You are in a physics lab working with a millimeter with a full scale reading of 50 milliamps. It contains a 0.1 ohm resistor in parallel with a 50 ohm galvanometer. You no need to use this instrument to measure voltage with a full scale reading of 50 volts and you cannot disassemble the A meter. How would you modify it to serve as a volt meter? And what would be its sensitivity in ohms per volt A says we can add a series resistor of 1000 ohms and modified instrument will be sensitive to 20 ohms per volt B says we can add a shunt resistor of 1500 ohms and it will be sensitive to 20 ohms per volt C says we can add a series resistor of 1000 ohms and it will be sensitive to 30 ohms per volt. And D says we can add a shunt resistor of 1500 ohms and it will be sensitive to 30 ohms per volt. Now, first of all, let's see if we can determine the total resistance of the ammeter. So we can better understand our problem. OK. Now, so far, we know that the AMET contains a 0.1 ohm resistor in parallel with a 50 ohm galvanometer. So that total resistance can be calculated by finding the total resistance for resistors in parallel. In other words, this tells us that the reciprocal of our total resistance is going to be equal to the sum of the reciprocal of the resistance of the emitter plus the resistance or the reciprocal of the resistance of the galvanometer. In this case, that's going to be equal to one divided by 0.1 ohms plus one divided by 50 ohms. And when we add the both of these, OK, then we get our reciprocal to be equal to 10.02 ohms. Therefore, that means our total resistance is the reciprocal of our result which is gonna be equal to 0.0998 ohms. Now that we have our total resistance, let's calculate the voltage drop across the ammeter at full scale. Now, recall in our problem, we are also told that the full scale current for the ammeter is 50 million amperes. OK? Or 0.05 amperes. So that tells us as well that the voltage across the ammeter can be calculated as the total current multiplied by or total resistance. So that's gonna be equal to 0.05 amperes multiplied by 0.00 sorry 0.0998 arms which gives us a voltage drop of 0.00499 volts. Now that we know the voltage drop across our emitter, then we can figure out the series resistor that we will need in this scenario. Now to measure a total voltage of 50 volts, the series resistor must drop the remaining voltage after accounting for the small voltage drop across the AM meter. OK. So that means the required voltage drop across the series. OK. First, we'll need to figure this out before we can calculate our series resistance. It is going to be equal to the voltage of the the galvanometer. 50 volts minus the voltage drop across the A meter. So that's 50 volts minus 0.00499 volts which equals 49.995 volts, which now means that the value of the series resistor is gonna be equal to the series voltage divided by the current 49.995 volts divided by 0.05 S which equals 999.9 ohms or to one significant figure 1000 ohms. Ok. No, this is the series resistor that we'll need. Ok, series resistor to to convert the millimeter to a volt into a volt meter. But now remember we had a second part to our problem. We need to figure out the sensitivity of the volt meter. No, the sensitivity is defined as the total resistance of the circuit divided by the full scale voltage, the total resistance is the sum of the ammeters resistance and the series resistor. OK. So now we also know then that the sensitivity is gonna be equal to the total resistance plus the series resistance divided by the full scale voltage. OK, which is gonna be 0.0998 ohms plus 999.9 ohms divided by 50 volts. Now, when we go ahead and answer here, then we should find that we get the sensitivity to be 20 ohms per vote. Therefore, the modified instrument will have a sensitivity of 20 ohms per foot. And we'll need to add a serious resistor of 1000 ohms to the circuit. Therefore, when we look back on our answer choices that tells us then that a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

(II) A milliammeter reads 25 mA full scale. It consists of a 0.20-Ω resistor in parallel with a 33-Ω galvanometer. How can you change this ammeter to a voltmeter giving a full scale reading of 25 V without taking the ammeter apart? What will be the sensitivity (Ω/V) of your voltmeter?

Key Concepts

Ammeter and Voltmeter Functionality

Parallel Resistors

Sensitivity of a Voltmeter

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