Discussion Forum

 A galvanometer has a resistance of 100 $\Omega$ and a current of 10 mA produces full-scale deflection in it. The resistance to being connected in series, to get a voltmeter of range 50 volt is 

Answer:

Explanation:

 A galvanometer is converted  into a voltmeter by connecting  a resistance in series with the galvanometer as shown in the circuit diagram.

 where, r is resistance of the resistor connected in series.

 Given galvanometer resistance ,$ R_{G}$  = 100 $\Omega$

 Voltmeter range V_max = 50 V and full deflection current l_fl  = 10 mA

 So, by applying the KVL in above  circuit diagram,

 $V_{AB}=100l_{fl}+Rl_{fl}$

 $\Rightarrow V_{AB}=(100+R)l_{fl}$

 $\therefore$    For a 50 V voltmeter  range there must be

    $V_{AB}$= 50 V  and  $l_{fl}$= 10 mA

 Now,  substituting the values of V_AB  and l_fl  in Eq.(i)

 we get,

 $50=(100+R)10\times10^{-3}$

$\therefore$    R=4900 $\Omega$

  Hence, the resistance to be connected in series to get a voltmeter of range 50 V is 4900 $\Omega$