Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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?

24 \Omega

36 \Omega

48 \Omega

60 \Omega

Step-by-Step Solution

Principle: The deflection ( $\phi$ ) in a moving coil galvanometer is directly proportional to the current flowing through its coil ( $I_g \propto \phi$ ) .
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: $I_g = I \left( \frac{S}{S+G} \right)$ .
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, $I_g \propto 10$ .
Calculation: Ratio of currents: $\frac{I_g}{I} = \frac{10}{50} = \frac{1}{5}$ . Substitute into the shunt formula: $\frac{1}{5} = \frac{12}{12 + G}$ . $12 + G = 12 \times 5$ $12 + G = 60$

$G = 48\ \Omega$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started