Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A moving coil galvanometer has a resistance of $50\ \Omega$ and gives full scale deflection for $10\ \text{mA}$ . How could it be converted into an ammeter with a full scale deflection for $1\ \text{A}$ ?

$50/99\ \Omega$ in series

$50/99\ \Omega$ in parallel

$0.01\ \Omega$ in series

$0.01\ \Omega$ in parallel

Step-by-Step Solution

Principle: To convert a galvanometer into an ammeter, a low resistance called a 'shunt' ( $S$ ) is connected in parallel with the galvanometer coil. This allows the excess current to bypass the galvanometer .
Formula: The potential difference across the galvanometer ( $G$ ) and the shunt ( $S$ ) is the same since they are in parallel. $I_g G = (I - I_g) S$ Where: $I_g$ = Full scale deflection current of galvanometer = $10\ \text{mA} = 0.01\ \text{A}$ $G$ = Resistance of galvanometer = $50\ \Omega$

$I$ = Desired range of ammeter = $1\ \text{A}$

Calculation: Rearranging for $S$ : $S = \frac{I_g G}{I - I_g}$ $S = \frac{0.01 \times 50}{1 - 0.01}$ $S = \frac{0.5}{0.99} = \frac{50}{99}\ \Omega$ .
Conclusion: A resistance of $50/99\ \Omega$ must be connected in parallel.

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