JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Current Electricity

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) $\frac{G}{200}$
(B) $\frac{G}{199}$
(C)199 G
(D)200 G

Solution

In an ammeter, the galvanometer is shunted by a low resistance S. Given that 5% of main current I passes through the galvanometer, 95% passes through the shunt.

Using current division:

I_{S} S = I_{g} G

Substituting:

\frac{95 I}{100} \cdot S = \frac{5 I}{100} \cdot G

Solving:

S = \frac{G}{19}

The ammeter resistance is:

R_{A} = \frac{S G}{S + G} = \frac{\frac{G ^{2}}{19}}{\frac{20 G}{19}} = \frac{G}{20}

Since

\frac{G}{20}

is not among the given options, this question was marked as Bonus.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option

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