The resistance required to be connected in parallel to an ammeter in order to increase its range 10 times, will be
A one-tenth the resistance of ammeter
B nine times the resistance of ammeter

C ten times the resistance of ammeter
D one-ninth the resistance of ammeter
The resistance required to be connected in parallel to an ammeter in order to increase its range 10 times, will be
A one-tenth the resistance of ammeter
B nine times the resistance of ammeter

C ten times the resistance of ammeter
D one-ninth the resistance of ammeter

Correct option is D one-ninth the resistance of ammeter
Full scale deflection voltage be V
Now initiall range is I

    \[\mathrm{V}=\mathrm{IR}\]

Now, Range is made 10 \mathrm{I}

    \[\begin{array}{l} \mathrm{V}=10 \mathrm{IR}^{\prime} \\ \mathrm{R}^{\prime}=\mathrm{V} / 10 \mathrm{I}=\mathrm{R} / 10 \end{array}\]

therefore, newR _{\underline{o q}} of parallel combination will be \mathrm{R} / 10 therefore,

    \[\begin{array}{l} 1 / R_{e q}=1 / R+1 / R_{p} \\ 1 / R_{p}=1 / R_{e q}-1 / R=10 / R-1 / R=9 / R \end{array}\]

therefore, \mathrm{R}_{\mathrm{p}}=\mathrm{R} / 9 option(D)