Question

A galvanometer has resistance '$G$' $\Omega$ and '$I_g$' is current flowing through it which produces full scale deflection. '$S_1$' is the value of shunt which converts it into an ammeter of range 0 to '$3I$' and '$S_2$' is the shunt value which converts it into an ammeter of range 0 to '$4I$', the ratio $S_2 : S_1$ is

• A. $\frac{4}{3}$
• B. $\frac{3I - I_g}{4I - I_g}$
• C. $\frac{3}{4}$
• D. $\frac{4I - I_g}{3I - I_g}$

Hint:

Shunt resistance is inversely proportional to the extra current that needs to bypass the galvanometer ($I_{\text{range}} - I_g$). A higher range ($4I$) naturally requires a smaller shunt ($S_2 < S_1$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given two different target ranges for an ammeter converted from the same galvanometer. We need to find the ratio of the two corresponding shunt resistances required.

Step 2: Key Formula or Approach:
To convert a galvanometer to an ammeter of range $I_{\text{range}}$, a shunt resistor $S$ is connected in parallel. The formula is:
$$S = \frac{I_g \times G}{I_{\text{range}} - I_g}$$

Step 3: Detailed Explanation:
For the first ammeter, the range is $3I$. The required shunt $S_1$ is:
$$S_1 = \frac{I_g G}{3I - I_g}$$
For the second ammeter, the range is $4I$. The required shunt $S_2$ is:
$$S_2 = \frac{I_g G}{4I - I_g}$$
We need the ratio $\frac{S_2}{S_1}$:
$$\frac{S_2}{S_1} = \frac{\left( \frac{I_g G}{4I - I_g} \right)}{\left( \frac{I_g G}{3I - I_g} \right)}$$
The term $I_g G$ cancels out from the numerator and denominator:
$$\frac{S_2}{S_1} = \frac{3I - I_g}{4I - I_g}$$

Step 4: Final Answer:
The ratio $S_2 : S_1$ is $\frac{3I - I_g}{4I - I_g}$, matching option (B).