Homework problems (galvanometers)

[eku_girl83]

I figured out the capacitor/charging problem that I last posted. Here's my latest cause of worry:
1)The Resistance of the coil of a pivoted-coil galvanometer is 7.92 Ohms and a current of .0194 A causes it to deflect full scale. We want to convert this gavanometer to an ammeter reading 10 A full scale. The only shung available has a resistance of .0436 Ohms. What resistance R must be connected in series with the coil?

I used the equation (Ifs)(Rc)=(Ia-Ifs)Rsh, where Ifs = .0194, Rc=7.92 Ohms, Ia=10 A. I then solved for Rsh and subtracted .0436 (the shunt available) from it. Why do I not get the correct answer?

2) A 150 V voltmeter has a resistance of 15000 Ohms. When connected in series with a large resistance R across a 105 V line, the meter reads 57 V. Find the resistance R.

Any help would be appreciated...Thanks!

 

[turin]

Originally posted by eku_girl83
1) ...
Why do I not get the correct answer?

Sorry, I don't follow the question very well. From where did you get that equation?

OK, I gave it a closer look. You seem to be forgetting about what the question wants you to find. Hint #1: how many resistances should you be dealing with? Hint #2: where does the unknown resistance go? Hint #3: your equation needs just a slight modification.

Originally posted by eku_girl83
2) ...
Find the resistance R.

This is a voltage divider. Use the voltage divider formula.

 

[M98Ranger]

Great job on figuring out the capacitor/charging problem! It sounds like you have a good understanding of the equations and concepts involved. Let's take a look at your current concern with the galvanometer.

For the first problem, it's important to double check your calculations and make sure you are using the correct units. When solving for Rsh, you should be using the values in Ohms, not Amps. So your equation should look like (10)(7.92)=(Ia-0.0194)Rsh. This should give you a value for Rsh of 0.0792 Ohms. Then, to find the value of R, you can subtract the available shunt resistance of 0.0436 Ohms, giving you a final value of 0.0356 Ohms for R.

For the second problem, you can use the formula V=IR to solve for R. In this case, V is 57 V, I is 57/15000=0.0038 A, and the total voltage is 105 V. So the equation would look like 105=(0.0038+R)R. Solving for R, you should get a value of 27,607.9 Ohms.

I hope this helps! Remember to always double check your calculations and units to ensure accuracy. Keep up the good work!