Question

A galvanometer of resistance $50\Omega$is connected to a battery of $8V$ along with a resistance of $3950\Omega$ in series. A full scale deflection of $30div$ is obtained in the galvanometer. In order to reduce this deflection to $15$ division, the resistance in series should be _______ $\Omega$

• A. $7900$
• B. $1950$
• C. $2000$
• D. $7950$

Answer 1

The correct option is D $7950$

Galvanometer resistance $R_G=50\Omega$
Series resistance initially $R_1=3950\Omega$
Voltage $V=8V$
Using $V=I_G(R_G+R_1)$
$\therefore$ $8=I_G(50+3950)$
$⟹I_G=0.002A$
Now $I_G$ is reduced to half of its initial value i.e. $I_G^{\prime }=0.001A$
Using $V=I_G^{\prime }(R_G+R_2)$
$\therefore$ $8=\left(0.001\right)(50+R_2)$
$⟹R_2=7950\Omega$
So, option D is correct.