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Finding the maximum value of current

  1. Jun 13, 2017 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    current sensitivity = $\frac{\theta}{I}$
    where $\theta$ is the division and $I$ is the current

    $I_{g}$ is the current through the galvanometer. $I$ is the total current. $G$ is galvanometer resistance and $S$ is the shunt resistance.
    3. The attempt at a solution

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


    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.
  2. jcsd
  3. Jun 13, 2017 #2
    What is current is passing through the meter without the shunt at full scale?
  4. Jun 13, 2017 #3
    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
  5. Jun 13, 2017 #4


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    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:
  6. Jun 14, 2017 #5
    Thanks for your help. I understood the mistake and got the correct solution.
  7. Jun 14, 2017 #6
    Your welcome
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