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: Understand the question.
A galvanometer with resistance $G$ gives full deflection at current $I_g$. Shunt $S_1$ makes it an ammeter reading up to $3I$, and shunt $S_2$ makes it read up to $4I$. We must find the ratio $S_2 : S_1$.

Step 2: Recall the shunt formula.
To turn a galvanometer into an ammeter of range $I_{\text{range}}$, we add a small shunt resistor in parallel. Its value is:
\[ S = \frac{I_g\, G}{I_{\text{range}} - I_g} \]

Step 3: Shunt for the first range.
For range $3I$:
\[ S_1 = \frac{I_g G}{3I - I_g} \]

Step 4: Shunt for the second range.
For range $4I$:
\[ S_2 = \frac{I_g G}{4I - I_g} \]

Step 5: Take the ratio.
Divide $S_2$ by $S_1$:
\[ \frac{S_2}{S_1} = \frac{\dfrac{I_g G}{4I - I_g}}{\dfrac{I_g G}{3I - I_g}} \]

Step 6: Cancel and simplify.
The $I_g G$ cancels from top and bottom, and the fractions flip:
\[ \frac{S_2}{S_1} = \frac{3I - I_g}{4I - I_g} \]
This matches option (2).
\[ \boxed{\frac{S_2}{S_1} = \frac{3I - I_g}{4I - I_g}} \]