Question

When a resistance of 12 $\Omega$ is connected parallel to a galvanometer, its pointer shift reduces from 50 marks to 20 marks. Determine the internal resistance of the galvanometer.

• A. 18 ohm
• B. 36 ohm
• C. 24 ohm
• D. 30 ohm

Answer 1

Answer: (A)

18 ohm

Answer 2

Let $G$ be the internal resistance of the galvanometer and $S$ be the shunt resistance. The deflection is proportional to the current through the galvanometer. When a shunt $S = 12 \Omega$ is connected in parallel, the galvanometer current $I_g'$ is related to the total current $I$ by $I_g' = I \frac{S}{G+S}$. The ratio of galvanometer currents is $\frac{I_g'}{I_g} = \frac{20}{50} = \frac{2}{5}$. Assuming the initial deflection of 50 marks was with the galvanometer alone, and the deflection of 20 marks is with the shunt $S=12\Omega$ connected in parallel. So, $\frac{I_g'}{I_g} = \frac{S}{G+S}$. Substituting the values: $\frac{2}{5} = \frac{12}{G+12}$ $2(G+12) = 5 \times 12$ $2G + 24 = 60$ $2G = 36$ $G = 18 \Omega$.