In order to pass 10% of the main current through a moving coil galvanometer of 99 ohms, the resistance of the

Question

In order to pass 10% of the main current through a moving coil galvanometer of 99 ohms, the resistance of the required shunt is:

Accepted Answer

1.  **Principle:** To convert a moving coil galvanometer into an ammeter, a low resistance called a 'shunt' ($S$) is connected in parallel with the galvanometer coil. This arrangement allows the majority of the current to bypass the sensitive galvanometer coil .
2.  **Parallel Connection:** Since the galvanometer ($G$) and the shunt ($S$) are in parallel, the potential difference across them is the same.
    $V_G = V_S \implies I_g G = (I - I_g) S$
    Where:
    *   $I$ is the total main current.
    *   $I_g$ is the current passing through the galvanometer.
    *   $I - I_g$ is the current passing through the shunt.
    *   $G$ is the resistance of the galvanometer.
3.  **Given Data:**
    *   Galvanometer Resistance $G = 99\ \Omega$.
    *   Current through galvanometer $I_g = 10\%$ of $I = 0.10 I$.
    *   Current through shunt $I_s = I - 0.10 I = 0.90 I$.
4.  **Calculation:**
    Substituting the values into the equation:
    $(0.10 I) 	imes 99 = (0.90 I) 	imes S$
    $S = \frac{0.10 	imes 99}{0.90}$
    $S = \frac{9.9}{0.9} = 11\ \Omega$.
5.  **Conclusion:** The required shunt resistance is $11\ \Omega$ .

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