A moving coil galvanometer has a resistance of $50\ \Omega$ and gives full scale deflection for $10\ ext{mA}

Question

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

Accepted Answer

1.  **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 .
2.  **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 = $10\ 	ext{mA} = 0.01\ 	ext{A}$
    *   $G$ = Resistance of galvanometer = $50\ \Omega$
    *   $I$ = Desired range of ammeter = $1\ 	ext{A}$
3.  **Calculation:**
    Rearranging for $S$:
    $S = \frac{I_g G}{I - I_g}$
    $S = \frac{0.01 	imes 50}{1 - 0.01}$
    $S = \frac{0.5}{0.99} = \frac{50}{99}\ \Omega$.
4.  **Conclusion:** A resistance of $50/99\ \Omega$ must be connected in parallel.

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