A galvanometer with an internal resistance of 100 ohm gives a full scale deflect

Question

A galvanometer with an internal resistance of 100 ohm gives a full scale deflection when the current is 50mA. This galvanometer is to be converted to a voltmeter with a range of 0 - 15V a) Determine the magnitude of the multiplying resistor necessary for successful conversion showing full working. b) This voltmeter is now connected as a part of the following circuit. Using the ACTUAL resistance of the voltmeter, calculate the combined resistance of the parallel components AB and hence the total resistance of the circuit. You may assume the battery has negligible the internal resistance. c). By calculating the voltage drop across AB. determine the current passing through the voltmeter. d) The current through a voltmeter is considered to be significant if the percentage of current through it is greater than 5% of total, Express the current calculated in (d) as a percentage of the total current through the circuit. Is the current through the voltmeter significant?

Explanation / Answer

To convert galvanometer in to volmeter attach a resistance in series with it value of it can be cal using formula

V =IgG+IgR
Solve it for R we wil get

V/Ig-G=R

15/50×10^-3-100=R=200ohm

1/200+1/40+1/60=1/R'

R'=21.28Ohm

So net resistance of circuit wil be 21.28+8=29.28ohm

Voltage drop across AB wil be

I=12÷29.28=0.41A

So voltage drop = 0.41×21.28=8.72 V

Current through volmeter wil be

8.72=I×200

I=0.044A

0.044/0.41×100=10.7%

Current through volmeter is significant

V

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