A galvanometer has an internal resistance of 32 Ω and deflects...

Question

A galvanometer has an internal resistance of 32 Ω and deflects full scale for a 48-μA current. Describe how to use this galvanometer to make an ammeter to read currents up to 25 A.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. You are in electronics lab and must measure currents up to 30 amps. The only available device is a galvanometer with a 50 ohms internal resistance which fully deflects at 60 micro amps. Your task is to modify this galvanometer to function as an AM editor that reads up to 30 amps. How would you accomplish this task? Awesome. So that's our goal. Our goal, what we're ultimately trying to solve for or ultimately what we're trying to do is we're trying to figure out how to modify this particular galvanometer in order for it to function as an ambi that can read up to 30 amps. And we're trying to figure out an explanation to how we would accomplish this specific task or this specific goal. So with that in mind, let's read off our multiple choice answers to see what our final answer might be a is con connect a shunt resistor of 1.0 multiplied by 10 to the power of negative four ohms in parallel with a galvanometer to convert the galvanometer into an emitter capable of reading up to 30 amps. B is connect a series resistor of 1.0 multiplied by 10 to the power negative four ohms in parallel with a galvanometer to convert the galvanometer into an ambi capable of reading up to 20 amps. C is connect a shunt resistor of 5.0 multiplied by 10 to the power negative four ohms in parallel with the galvanometer to convert the galvanometer into an emitter capable of reading up to 30 amps. And finally, for D connect a series resistor of 5.0 multiplied by 10 to the power of negative four ohms in parallel with the galvanometer to convert the galvanometer into an amid capable of reading up to 20 amps. OK. So first off, we need to determine the voltage across the galvanometer at the full scale deflection or rather when the full scale deflection takes place. So we need to recall and use the equation to help us solve for the voltage across the galvanometer, which will call this value V subscript G. And we can determine this value by recalling and using ohm's law. So VG which is the voltage across the galvanometer is equal to the current across the galbo meter multiplied by the resistance across the galbo meter. And we need to note really quick off to the side here in blue, let's make a little note here that IG in this particular case will be equal to 60 micro amps, which is also equal to 60 multiplied by 10 to the power of negative six amps. I've gone ahead and performed the conversion from micro amps to amps for you already. But you can use dimensional analysis to perform this conversion or you just quickly look it up. But to save some time, I just gave it to you. We also know that RG, the resistance across the galvanometer is equal to 50 ohms. Awesome. So now all we have to do is plug in our known values to solve for the numerical value of VG and VG is equal to 60 multiplied by 10 to the power of negative six amps. And the reason why we had to convert micro amps to amps is to make sure that we get our final units for VG and volts. So now we need to multiply our current by the resistance which our resistance value is 50 ohms. So when we plug that into our calculator, we will get 0.003 volts, which is also equal to when we write it in scientific notation, three, multiplied by 10 to the power of negative three volts. Awesome. So now our second step is we need to calculate the total current I that the aim should read. We also need to take into account that the desired current range for the modified animator is. And let's write this important note in blue. So we need to note that the desired current current range for this modified amid that we're dealing with will be I equals 30 amps. And this is given to us in the problem itself. So with that in mind, we need to now figure out what the shunt resistor, which we're gonna deer denote the shunt resistor as RS. So figuring out what the shunt resistor is, we need to recall and note that the shunt resistor must carry the difference between the total current I and the galvanometer. So galvanometer current IG which we can write this equation as, as we should recall, that is. So the current across the shunt resistor is equal to I where I is the total current minus IG where IG is the galvanometer current. So plugging in our known variables will give us 30 amps for I. So we take 30 amps for I and then we need to subtract 60 multiplied by 10 to the power of negative six amps for IG is that's our IG value. So when we plug that into our calculator, and we round to the nearest whole number, we will find that it's approximately equal to 30 amps, which is awesome is when you initially type this into your calculator, you'll get 29.999 amps, which when we round to the nearest whole number will be 30 amps, which is fantastic. So now at this point, we need to determine the voltage across the shunt resistor which we'll call vs. And we need to note that this is the same as the voltage across the galvanometer. So as we should recall, we could write that vs which is the voltage across the shunt resistor is equal to VG, which is all equal to three, multiplied by 10 to the power of negative three volts. So once again, when we recall and use Ohm Slaw, we can calculate the shunt resistors value. So we could solve for RS. So the resistance of the shunt resistor is equal to vs divided by is which plugging in are known variables. VS is equal to three multiplied by 10 to the power of negative three faults. And we divide this by is which is 30 amps. So let me plug this into our calculator. We will get 1.0 multiplied by 10 to the power of negative four ohms. And that's it. That's our final answer when rounded to one decimal place hooray, we did it. So in order to modify this galvanometer to function as an emitter that reads up to 30 amps, we will need to connect a shunt resistor in parallel with the galvanometer. Therefore, if we, so if we connect a shunt resistor of 1.0 multiplied by 10 to the power of negative four ohms in parallel with the galvanometer, we will be able to convert this particular galvanometer into an emitter that is capable of reading up to 30 amps. So that means without a doubt, the correct answer has to be the letter A which once again is connect, we need to connect a shunt resistor of 1.0 multiplied by 10 to the power of negative four ohms in parallel with a galvanometer to convert this galvanometer into an amid capable of reading up to 30 amps. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A galvanometer has an internal resistance of 32 Ω and deflects full scale for a 48-μA current. Describe how to use this galvanometer to make an ammeter to read currents up to 25 A.

Key Concepts

Galvanometer

Shunt Resistor

Current Division

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