- #1

Muthumanimaran

- 81

- 2

## Homework Statement

galvanometer with 50 divisions on the scale requires a current sensitivity of 0.1 m A/division. The resistance of the galvanometer is 40Ω. If a shunt resistance 0.1 Ω is connected across it, find the maximum value of the current that can be measured using this ammeter.

## Homework Equations

current sensitivity = $\frac{\theta}{I}$

where $\theta$ is the division and $I$ is the current

$\frac{I_{g}}{I-{I_{g}}}G=S$

$I_{g}$ is the current through the galvanometer. $I$ is the total current. $G$ is galvanometer resistance and $S$ is the shunt resistance.

## The Attempt at a Solution

using the first expression, I found the current through the circuit, i.e,

$$I=\frac{50}{0.1mA}$$

or

$$I=5\times10^{5}$$

next I substituted in the above formula, I get $I_{g}$ is equal to 1247 A. But it is not the correct answer. The correct answer is 2 A. Definitely I made a mistake, I understand the mistake is purely conceptual. I believe the maximum current that can be measured is not $I_{g}$, so help me to understand the problem.