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

Answer 1

A high resistance that is connected in series with the galvanometer to convert it into voltmeter. The value of the resistance is given by ${\displaystyle R=\frac{V}{I_g}-G}$

Here,
V = Potential difference across the terminals of the voltmeter

I_g = Current through the galvanometer

G = Resistance of the galvanometer

When resistance R₁ is connected in series with the galvanometer,

${\displaystyle R_1=\frac{V}{I_g}-G}$

When resistance R₂ is connected in series with the galvanometer,

 ${\displaystyle R_2=\frac{V}{2I_g}-G\text{.....(ii)}}$

From (i) and (i), we get

${\displaystyle \frac{V}{2I_g}=R_1-R_2\text{and}G=R_1-2R_2}$

The resistance R₃ required to convert the given galvanometer into voltmeter of range 0 to 2V is given by

${\displaystyle R_3=\frac{2V}{I_g}-G}$

⇒R₃=4(R₁−R₂)−(2R₁−R₂)=3R₁−2R₂

G in terms of R₁ and R₂ is given by

G=R₁−2R₂