Galvanometer problems

  • Thread starter eku_girl83
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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!
 

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  • #2
Integral
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The addition resistance that you are looking for is in series with the COIL not the shunt. So replace RC in your expression with a resistance equivalent to the one you need to balance the Right hand side. so instead of RC use:
Re=RC+Rs

Where Rs is the additional series resistance.
 
  • #3
Integral
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2:

You know that the meter reads 57V with a source of 105V, this means that 105V-57V is dropped across the internal resistance of the meter. You know the internal resistance so can compute the current. you now now voltage and current (series circuit so the current is equal) in the external resistance. Apply Ohms law again to get the external resistance.
 

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