When a 12 \Omega resistor is connected in parallel with a moving coil galvanometer, its deflection reduces fro

Question

When a 12 \Omega resistor is connected in parallel with a moving coil galvanometer, its deflection reduces from 50 divisions to 10 divisions. What will be the resistance of the galvanometer?

Accepted Answer

1.  **Principle:** The deflection ($\phi$) in a moving coil galvanometer is directly proportional to the current flowing through its coil ($I_g \propto \phi$) .
2.  **Current Divider Rule:** When a shunt resistance ($S$) is connected in parallel with the galvanometer (resistance $G$), the main current ($I$) divides between them. The current flowing through the galvanometer ($I_g$) is given by: $I_g = I \left( \frac{S}{S+G} ight)$.
3.  **Data Interpretation:**
    *   The initial deflection corresponds to the full current $I$ passing through the galvanometer (before shunting, or if the shunt wasn't diverting current). Let deflection $\phi_1 = 50$. So, $I \propto 50$.
    *   After connecting the shunt $S = 12\ \Omega$, the deflection reduces to $\phi_2 = 10$. This deflection corresponds to the new current $I_g$ flowing through the galvanometer. So, $I_g \propto 10$.
4.  **Calculation:**
    *   Ratio of currents: $\frac{I_g}{I} = \frac{10}{50} = \frac{1}{5}$.
    *   Substitute into the shunt formula: $\frac{1}{5} = \frac{12}{12 + G}$.
    *   $12 + G = 12 	imes 5$
    *   $12 + G = 60$
    *   $G = 48\ \Omega$.

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