Finding the maximum value of current

[Muthumanimaran]

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.

 

[Inventive]

What is current is passing through the meter without the shunt at full scale?

 

[Inventive]

The reason I ask is because if full scale is 50 divisions and there is .1ma (100 microamps) per division then the meter will draw 5 ma at full scale. Knowing this in conjunction with the 40 ohm series resistance of the meter means it has a voltage of 40 x 5 ma or .2 volts. Now the .1 ohm is in parallel with the 40 ohm series meter resistance. At this point the current through the .1 ohm resistance can be found and from that the maximum current that will drive the meter to full scale. Let me know if you still need help

 

[haruspex]

current sensitivity of 0.1 m A/division.
...
current sensitivity = $\frac{\theta}{I}$ where $\theta$ is the division and $I$ is the current

Do you not see a conflict between those two statements?

In this forum, the latex requires either a double dollar sign (giving it a line to itself) or a double hash sign (#) to embed it within a line:

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

 

[Muthumanimaran]

The reason I ask is because if full scale is 50 divisions and there is .1ma (100 microamps) per division then the meter will draw 5 ma at full scale. Knowing this in conjunction with the 40 ohm series resistance of the meter means it has a voltage of 40 x 5 ma or .2 volts. Now the .1 ohm is in parallel with the 40 ohm series meter resistance. At this point the current through the .1 ohm resistance can be found and from that the maximum current that will drive the meter to full scale. Let me know if you still need help

Thanks for your help. I understood the mistake and got the correct solution.

 

[Inventive]

Your welcome