Question

A galvanometer of resistance $50\,\Omega$ is connected to a battery of $8\,{\text{V}}$ along with a resistance of $3950\,\Omega$ in series. A full scale deflection of $30\,{\text{div}}$ is obtained in the galvanometer. In order to reduce this deflection to 15 divisions, the resistance in series should be - - - $\Omega$.
A. 7900
B. 1950
C. 2000
D. 7950

Answer 1

Use the expression for Ohm’s law. Using the expression for Ohm’s law, determine the current for which the galvanometer shows maximum deflection of 30 divisions. Using this value of the galvanometer current, determine the value of the resistance that should be connected in series using Ohm’s law expression in order to have the deflection of 15 divisions.

Formula used:
The expression for Ohm’s law is
$V = IR$ …… (1)
Here, $V$ is the potential difference between the ends of the conductor, $I$ is the current flowing in the conductor and $R$ is resistance of the conductor.

Complete step by step answer:
We have given that the resistance of the galvanometer is $50\,\Omega$ and the resistance of the resistor connected in series is $3950\,\Omega$.
$G = 50\,\Omega$
$R = 3950\,\Omega$
The galvanometer is connected to a battery of $8\,{\text{V}}$.
$V = 8\,{\text{V}}$
The full scale deflection obtained in the galvanometer is $30\,{\text{div}}$.We know that the electric current through the resistors connected in series is the same.

Let us first determine the current for which the galvanometer shows maximum deflection.According to Ohm’s law, the potential difference for the system of galvanometer and the resistance connected in series with the galvanometer is given by
$V = {I_G}\left( {G + {R_1}} \right)$
Here, ${I_G}$ is the current for which the galvanometer shows maximum deflection.
Substitute $8\,{\text{V}}$ for $V$, $50\,\Omega$ for $G$ and $3950\,\Omega$ for ${R_1}$ in the above equation.
$8\,{\text{V}} = {I_G}\left( {50\,\Omega + 3950\,\Omega } \right)$
$\Rightarrow 8\,{\text{V}} = {I_G}\left( {4000\,\Omega } \right)$
$\Rightarrow {I_G} = \dfrac{{8\,{\text{V}}}}{{4000\,\Omega }}$
$\Rightarrow {I_G} = 0.002\,{\text{A}}$
Hence, the current for which the galvanometer shows maximum deflection is $0.002\,{\text{A}}$.

Now we have to determine the value of the resistance ${R_2}$ in series for which the full scale deflection of the galvanometer is reduced to half of its initial value.Hence, the current for which the galvanometer shows the maximum current becomes
$I_G^1 = \dfrac{{{I_G}}}{2}$
Thus, the potential difference across the galvanometer and resistor in series is given by
$V = I_G^1\left( {G + {R_2}} \right)$
Substitute $\dfrac{{{I_G}}}{2}$ for $I_G^1$ in the above equation.
$V = \dfrac{{{I_G}}}{2}\left( {G + {R_2}} \right)$
Substitute $8\,{\text{V}}$ for $V$, $50\,\Omega$ for $G$ and $0.002\,{\text{A}}$ for ${I_G}$ in the above equation.
$8\,{\text{V}} = \dfrac{{0.002\,{\text{A}}}}{2}\left( {50\,\Omega + {R_2}} \right)$
$\Rightarrow 8 = 0.001\left( {50 + {R_2}} \right)$
$\Rightarrow {R_2} = 8000 - 50$
$\therefore {R_2} = 7950\,\Omega$
Therefore, the resistance that should be connected in series is $7950\,\Omega$.

Hence, the correct option is D.

Note: The students should keep in mind that the other resistance is connected in series with the galvanometer. Hence, the net resistance of the galvanometer and the other resistance is taken as addition of these two resistance. If this other resistance is connected in parallel then we have to use the formula for equivalent resistance of two resistors connected in parallel.