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MHTCET 10 May 2024 Shift 2 Paper

Question : 48 of 150

Marks: +1, -0

In an ammeter,

0.25 \%

of main current passes through the galvanometer. If the resistance of the galvanometer is '

G

', the resistance of ammeter will be

$\frac{399}{400} \mathrm{G}$
$\frac{1}{400} \mathrm{G}$
$\frac{499}{500} \mathrm{G}$
$\frac{1}{500} \mathrm{G}$

Validate

Solution:

Concept:

An ammeter uses a small shunt resistance in parallel with the galvanometer. The effective ammeter resistance is the parallel combination of the galvanometer resistance

G

and the shunt resistance

S

Explanation:

Let the total main current be

I

. Only

0.25\%

I

flows through the galvanometer, so

I_g = \frac{0.25}{100}I = \frac{I}{400}

The remaining current flows through the shunt:

I_s = I - I_g = I - \frac{I}{400} = \frac{399}{400}I

Since the galvanometer and shunt are in parallel, the potential difference across both is equal:

I_g G = I_s S

Substituting the currents:

\frac{I}{400}G = \frac{399}{400}I \cdot S

, which gives

S = \frac{G}{399}

The ammeter resistance is the parallel combination of

G

and

S

R = \frac{GS}{G+S}

Putting

S = \frac{G}{399}

R = \frac{G \cdot \frac{G}{399}}{G + \frac{G}{399}} = \frac{\frac{G^2}{399}}{\frac{400G}{399}} = \frac{G}{400}

Answer:

The resistance of the ammeter is

\frac{1}{400}G

. Therefore, the correct option is B.

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