Q140 / 180 · COMEDK 2024 (Morning Shift) · Physics
The resistance of the galvanometer and shunt of an ammeter are $90 \mathrm{~ohm}$ and $10 \mathrm{~ohm}$ respectively, then the fraction of the main current passing through the galvanometer and the shut respectively are:
- A$\dfrac{1}{10}$ and $\dfrac{1}{90}$
- Correct answerB$\dfrac{1}{10}$ and $\dfrac{9}{10}$
- C$\dfrac{9}{10}$ and $\dfrac{1}{10}$
- D$\dfrac{1}{90}$ and $\dfrac{1}{10}$
Solution
Let $I$ be the total current, $I_g$ be the current through the galvanometer, and $I_s$ be the current through the shunt.
Given the resistance of the galvanometer $G = 90 \, \Omega$ and the resistance of the shunt $S = 10 \, \Omega$.
Since the galvanometer and shunt are connected in parallel, the potential difference across them is the same:
$I_g G = I_s S$
$I_g (90) = I_s (10)$
$I_s = 9 I_g$
The total current $I$ is given by $I = I_g + I_s$. Substituting $I_s = 9 I_g$ into this equation:
$I = I_g + 9 I_g = 10 I_g$
The fraction of the main current passing through the galvanometer is $\dfrac{I_g}{I} = \dfrac{I_g}{10 I_g} = \dfrac{1}{10}$.
The fraction of the main current passing through the shunt is $\dfrac{I_s}{I} = \dfrac{9 I_g}{10 I_g} = \dfrac{9}{10}$.
Answer: $\dfrac{1}{10}$ and $\dfrac{9}{10}$
Given the resistance of the galvanometer $G = 90 \, \Omega$ and the resistance of the shunt $S = 10 \, \Omega$.
Since the galvanometer and shunt are connected in parallel, the potential difference across them is the same:
$I_g G = I_s S$
$I_g (90) = I_s (10)$
$I_s = 9 I_g$
The total current $I$ is given by $I = I_g + I_s$. Substituting $I_s = 9 I_g$ into this equation:
$I = I_g + 9 I_g = 10 I_g$
The fraction of the main current passing through the galvanometer is $\dfrac{I_g}{I} = \dfrac{I_g}{10 I_g} = \dfrac{1}{10}$.
The fraction of the main current passing through the shunt is $\dfrac{I_s}{I} = \dfrac{9 I_g}{10 I_g} = \dfrac{9}{10}$.
Answer: $\dfrac{1}{10}$ and $\dfrac{9}{10}$
Hence, the correct answer is option: 'B'.