Q109 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
A galvanometer has a current range of $10$ mA and a voltage range of $0.75$ V. To convert this galvanometer into an ammeter of range $10$ A, what is the shunt resistance?
- A$\dfrac{100}{999} \Omega$
- B$\dfrac{50}{999} \Omega$
- C$\dfrac{200}{999} \Omega$
- Correct answerD$\dfrac{75}{999} \Omega$
Solution
Given $I_g = 10 \text{ mA} = 0.01 \text{ A}$ and $V_g = 0.75 \text{ V}$.
Resistance of the galvanometer is $G = \dfrac{V_g}{I_g} = \dfrac{0.75}{0.01} = 75 \, \Omega$.
To convert the galvanometer into an ammeter of range $I = 10 \text{ A}$, a shunt resistance $S$ is connected in parallel.
The value of the shunt resistance is given by $S = \dfrac{I_g G}{I - I_g} = \dfrac{V_g}{I - I_g}$.
Substituting the values, we get $S = \dfrac{0.75}{10 - 0.01} = \dfrac{0.75}{9.99} = \dfrac{75}{999} \, \Omega$.
Answer: $\dfrac{75}{999} \Omega$
Resistance of the galvanometer is $G = \dfrac{V_g}{I_g} = \dfrac{0.75}{0.01} = 75 \, \Omega$.
To convert the galvanometer into an ammeter of range $I = 10 \text{ A}$, a shunt resistance $S$ is connected in parallel.
The value of the shunt resistance is given by $S = \dfrac{I_g G}{I - I_g} = \dfrac{V_g}{I - I_g}$.
Substituting the values, we get $S = \dfrac{0.75}{10 - 0.01} = \dfrac{0.75}{9.99} = \dfrac{75}{999} \, \Omega$.
Answer: $\dfrac{75}{999} \Omega$
Hence, the correct answer is option: 'D'.