Question

A galvanometer of resistance of $50\Omega$ is connected to a battery of 3V along with a resistance of $2950\Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions the resistance in series should be connected whose value is

Answer 1

In the above question we are basically constructing a voltmeter. We are asked to determine the resistance in series that is to be connected to the above set up such that the deflection for 3V connection across it corresponds to 20 divisions. Hence we will use Kirchhoff’s law of voltage and the ohm’s law to determine the resistance required in series such that the above condition is satisfied.

Formula used:
$V={{V}_{1}}+{{V}_{2}}+.......+{{V}_{n}}$
$V=IR$

Complete step by step answer:
Let us say we connect a battery across n resistances in series such that the potential difference across them is V. If ${{V}_{1}},{{V}_{2}},.......{{V}_{n}}$ represents the individual potential difference across them, then by Kirchhoff’s law of voltage we get,
$V={{V}_{1}}+{{V}_{2}}+.......+{{V}_{n}}$
Let us say for the same is the effective resistance in series is R, then current I in the circuit is given by ohms law i.e. $V=IR$
In the above question we are given that a galvanometer of resistance of G=$50\Omega$ is connected to a battery of 3V along with a resistance of r=$2950\Omega$ in series.
Hence the current (i) in the circuit is equal to,
$\begin{aligned} & V={{V}_{r}}+{{V}_{G}} \\\ & \Rightarrow V=i(r+G) \\\ & \Rightarrow i=\dfrac{V}{r+G}...(1) \\\ & \Rightarrow i=\dfrac{3}{2950+50}={{10}^{-3}}A \\\ \end{aligned}$
Now for this value of current the galvanometer basically gives a full scale deflection of 30 divisions. Hence we can imply that the galvanometer will show a deflection of 20 division when the current (${{I}_{G}}$ )in the circuit is,
$\begin{aligned} & {{I}_{G}}=\dfrac{20}{30}i \\\ & \Rightarrow {{I}_{G}}=\dfrac{2}{3}{{10}^{-3}}A=0.66\times {{10}^{-3}}A \\\ \end{aligned}$
Using equation 1 the for (${{I}_{G}}$) value of current the resistance R to be connected in series is equal to,
$\begin{aligned} & {{I}_{G}}=\dfrac{V}{r+G} \\\ & \Rightarrow \dfrac{2}{3}\times {{10}^{-3}}=\dfrac{3}{R+50} \\\ & \Rightarrow R+50=\dfrac{9}{2}\times {{10}^{3}}=4500 \\\ & \therefore R=4500-50=4450\Omega \\\ \end{aligned}$
Hence the required value of resistance is 4450 ohms.

Note:
It is to be noted that the deflection given by the galvanometer corresponds to the degree of current through it. In the above case by connecting a very high value of resistance we basically try to construct a voltmeter. This is because by connecting very high values of resistance we ensure that the galvanometer draws minimum current.