Question

Current sensitivity of a moving coil galvanometer is $5\left(\frac{\text{div}}{\text{mA}}\right)$ and its voltage sensitivity (angular deflection per unit voltage applied) is $20\left(\frac{\text{div}}{\text{V}}\right).$ The resistance of the galvanometer is-

• A. $150\Omega$
• B. $200\Omega$
• C. $250\Omega$
• D. $300\Omega$

Answer 1

The correct option is C $250\Omega$

Given,
Current sensitivity $I_s$ is,

$I_s=\frac{\theta }{L}=\frac{NBA}{C}=5\left(\frac{\text{div}}{\text{mA}}\right)$

$V_s=\frac{\theta }{V}=\frac{NBA}{C{R}_g}=20\left(\frac{\text{div}}{\text{V}}\right)$

Where,
$I_s=\text{Current sensitivity};V_s=\text{Voltage sensitivity};N=\text{Number of turns}$

$B=\text{Magnetic field};A=\text{Area};C=\text{Couple per unit twist}$

Resistance of the moving coil galvanometer is,

$R_g$= $\frac{I_s}{V_s}$= $\frac{5}{20\times {10}^{-3}}$ = $\frac{5000}{20}$

$R_g$= $250\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.