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A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R_{1} in series with the coil. If a resistance R_{2} is connected in series with it, then it can measures upto V/2 volts. Find the resistance, in terms of R_{1} and R_{2}, 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 R_{1} and R_{2}

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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

I_{g} = Current through the galvanometer

G = Resistance of the galvanometer

When resistance R_{1} is connected in series with the galvanometer,

`R_1=V/I_g-G`

When resistance R_{2} 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 R_{3} required to convert the given galvanometer into voltmeter of range 0 to 2V is given by

`R_3=(2V)/I_g-G`

⇒R_{3}=4(R_{1}−R_{2})−(2R_{1}−R_{2})=3R_{1}−2R_{2}

G in terms of R_{1} and R_{2} is given by

G=R_{1}−2R_{2}

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