Discussion Forum

 A galvanometer gives full-scale deflection with 0.006 A current. By connecting it to a 4990Ω resistance, it can be converted into a voltmeter of range 0-30V. If connected to a  $\frac{2n}{249}$ Ω resistance, it becomes an ammeter of range 0-1.5 A. The value of n is 

Answer:

Explanation:

$i_{g}(G+4990)=V$

$\Rightarrow$     $\frac{6}{1000}(G+4990)=30$

$\Rightarrow$    $G+4990=\frac{30000}{6}=5000$

$\Rightarrow$             G=10Ω

                       V_ab  =V_cd

$\Rightarrow$               $i_{g}G=(1.5-i_{g})S$

 $\Rightarrow$      $\frac{6}{1000}\times10 = \left( 1.5-\frac{6}{1000}\right)S$

    $\Rightarrow$    $S=\frac{60}{1494}=\frac{2n}{249}$

$\Rightarrow$      $n=\frac{249\times30}{1494}=\frac{2490}{498}=5$