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Question
A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measures upto V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2
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Solution
A high resistance that is connected in series with the galvanometer to convert it into voltmeter. The value of the resistance is given by `R=V/I_g-G`
Here,
V = Potential difference across the terminals of the voltmeter
Ig = Current through the galvanometer
G = Resistance of the galvanometer
When resistance R1 is connected in series with the galvanometer,
`R_1=V/I_g-G`
When resistance R2 is connected in series with the galvanometer,
`R_2=V/(2I_g)-G" .....(ii)"`
From (i) and (i), we get
`V/(2I_g)=R_1-R_2 " and "G=R_1-2R_2`
The resistance R3 required to convert the given galvanometer into voltmeter of range 0 to 2V is given by
`R_3=(2V)/I_g-G`
⇒R3=4(R1−R2)−(2R1−R2)=3R1−2R2
G in terms of R1 and R2 is given by
G=R1−2R2
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