Question

A galvanometer of resistance $25\Omega$ is connected to a battery of $2V$ along with a resistance in series. When the value of this resistance is $3000\Omega$ a full scale deflection of $30$ units is obtained in the galvanometer. In order to reduce this deflection to $20$ units, the resistance in series will be:

• A. $1991.66\Omega$
• B. $5413\Omega$
• C. $2000\Omega$
• D. $6000\Omega$

Answer 1

The correct option is A $1991.66\Omega$

$V\alpha$ deflection

$\frac{V_1}{V_2}=\frac{30}{20}=\frac{R_1}{R_2}$

$R_2=\frac{2}{3}(25+3000)$

$=2016.66\Omega$

$R=R_2-G$

$=(2016.66-25)\Omega$

$=1991.66\Omega$