Troubleshooting Galvanometer Problems and Calculating Resistance | Expert Tips

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SUMMARY

The discussion focuses on troubleshooting galvanometer problems and calculating necessary resistances for conversion to an ammeter. The resistance of the galvanometer coil is 7.92 Ohms, and the goal is to modify it for a full-scale reading of 10 A using a shunt with a resistance of 0.0436 Ohms. The equation used is (Ifs)(Rc)=(Ia-Ifs)Rsh, where Ifs is 0.0194 A and Ia is 10 A. Additionally, a problem involving a 150 V voltmeter with a resistance of 15,000 Ohms connected to a 105 V line is discussed, requiring the calculation of an external resistance R based on the voltmeter's reading of 57 V.

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  • Understanding of galvanometer operation and conversion to ammeter
  • Familiarity with Ohm's Law and series circuits
  • Knowledge of shunt resistor calculations
  • Ability to manipulate electrical equations for circuit analysis
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  • Study the principles of galvanometer to ammeter conversion
  • Learn about shunt resistor design and calculations
  • Explore advanced applications of Ohm's Law in circuit analysis
  • Investigate the effects of internal resistance in measurement devices
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Electrical engineers, technicians working with measurement instruments, and students studying circuit theory will benefit from this discussion.

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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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.
 
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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