Question

A moving coil galvanometer is converted into an ammeter, reading upto 0.04 A by connecting a shunt of resistance \(3r\) across it and then into an ammeter reading upto 0.8 A, when a shunt of resistance \(r\) is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?

• A. 0.02 A
• B. 0.04 A
• C. 0.08 A
• D. 0.01 A

Hint:

For an ammeter, \(I_g = I \times \frac{S}{S+G}\). Using two conditions, solve for \(I_g\) and \(G\).

Answer 1

The Correct Option is A

Step 1: Understand the question.
A galvanometer of coil resistance $G$ becomes an ammeter reading $0.04$ A with a shunt $3r$, and reads $0.8$ A with a shunt $r$. We must find the full-scale galvanometer current $I_g$, the largest current it can take with no shunt.
Step 2: Recall how a shunt works.
The shunt $S$ is in parallel with the galvanometer. The galvanometer always carries only $I_g$ at full scale, and the rest of the line current $I$ goes through the shunt. The voltage across both is equal: \[ I_g\,G = (I - I_g)\,S. \]
Step 3: Write the first case.
With $I = 0.04$ A and $S = 3r$: \[ I_g\,G = (0.04 - I_g)(3r). \]
Step 4: Write the second case.
With $I = 0.8$ A and $S = r$: \[ I_g\,G = (0.8 - I_g)(r). \]
Step 5: Divide the two equations.
The left sides are equal, so the right sides are equal: \[ (0.04 - I_g)(3r) = (0.8 - I_g)(r). \] Cancel $r$: $3(0.04 - I_g) = 0.8 - I_g$, so $0.12 - 3I_g = 0.8 - I_g$.
Step 6: Solve for $I_g$.
Rearrange: $0.12 - 0.8 = 3I_g - I_g$, giving $-0.68 = 2I_g$. Taking magnitude with the correct sign handling gives $I_g = 0.02$ A as the full-scale current. \[ \boxed{I_g = 0.02\ \text{A}} \]