Question 327 of 380

In an ammeter, $5$ of the main current passes through the galvanometer. If resistance of the galvanometer is $G,$ the resistance of ammeter will be:

A$\dfrac{G}{20}$B$\dfrac{G}{199}$C$199G$D$200G$

Show answer and solution▾

Answer: (A) $\dfrac{G}{20}$

As potential drop across both branches would be the same, we get ${I}_{S}S={I}_{g}G$
$\left(\dfrac{95}{100}I\right)S=\left(\dfrac{5I}{100}\right)G$
$\Rightarrow S=\dfrac{G}{19}$
Now, resistance of ammeter will be
${R}_{A}=\dfrac{SG}{S+G}=\dfrac{\dfrac{{G}^{2}}{19}}{\dfrac{20G}{19}}$
$\Rightarrow {R}_{A}=\dfrac{G}{20}$

JEE Main Physics practice question on Current Electricity (Electric Instruments), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.