Q170 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
A galvanometer of resistance $50 \Omega$ is connected to a battery of 4 V along with a resistance of $3950 \Omega$ in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 10 divisions, the resistance in series should be equal to:
- A6000 ohm
- B8950 ohm
- C7000 ohm
- Correct answerD11950 ohm
Solution
The current $I$ flowing through the galvanometer for a full-scale deflection of $30$ divisions is given by Ohm's law:
$I = \dfrac{V}{R_{g} + R_{s}} = \dfrac{4}{50 + 3950} = \dfrac{4}{4000} = 10^{-3} \text{ A}$
The current per division is $\dfrac{10^{-3}}{30} \text{ A/division}$.
To obtain a deflection of $10$ divisions, the new current $I'$ required is:
$I' = 10 \times \dfrac{10^{-3}}{30} = \dfrac{1}{3} \times 10^{-3} \text{ A}$
Let the new series resistance be $R'_{s}$. The circuit equation is:
$I' = \dfrac{V}{R_{g} + R'_{s}}$
$\dfrac{1}{3} \times 10^{-3} = \dfrac{4}{50 + R'_{s}}$
$50 + R'_{s} = 4 \times 3 \times 10^{3} = 12000$
$R'_{s} = 12000 - 50 = 11950 \Omega$
Answer: 11950 ohm
$I = \dfrac{V}{R_{g} + R_{s}} = \dfrac{4}{50 + 3950} = \dfrac{4}{4000} = 10^{-3} \text{ A}$
The current per division is $\dfrac{10^{-3}}{30} \text{ A/division}$.
To obtain a deflection of $10$ divisions, the new current $I'$ required is:
$I' = 10 \times \dfrac{10^{-3}}{30} = \dfrac{1}{3} \times 10^{-3} \text{ A}$
Let the new series resistance be $R'_{s}$. The circuit equation is:
$I' = \dfrac{V}{R_{g} + R'_{s}}$
$\dfrac{1}{3} \times 10^{-3} = \dfrac{4}{50 + R'_{s}}$
$50 + R'_{s} = 4 \times 3 \times 10^{3} = 12000$
$R'_{s} = 12000 - 50 = 11950 \Omega$
Answer: 11950 ohm
Hence, the correct answer is option: 'D'.