Exam Tomorrow! Can someone help?? Stuck!!

[AnthonyR23]

Hi, I am working on a unit for my electro- course... I have the answer but am having trouble working out this question.... if someone could help that would be awesome!!

Question:
If a Weston-type ammeter with a coil resistance of 12.6 ohms has a shunt resistor of 1.92 ohms, which is dissipating 930 mW of power, how much current is going through the meter coil?

Answer:
106.1 mA

I am trying to work this out... but am having a lot of trouble...

Formula Shunt Current / Meter Current = Meter Resistance / Shunt Resistance

Thanks!! This is the one question that is holding me up... before my exam...

 

[Roff]

Hi, I am working on a unit for my electro- course... I have the answer but am having trouble working out this question.... if someone could help that would be awesome!!

Question:
If a Weston-type ammeter with a coil resistance of 12.6 ohms has a shunt resistor of 1.92 ohms, which is dissipating 930 mW of power, how much current is going through the meter coil?

Answer:
106.1 mA

I am trying to work this out... but am having a lot of trouble...

Formula Shunt Current / Meter Current = Meter Resistance / Shunt Resistance

Thanks!! This is the one question that is holding me up... before my exam...

The resistor power dissipation is P=V²/R, where V is the voltage across the resistor.
Therefore, V=√(P×R).
This is also the voltage across the meter, since it is in parallel with the resistor.
The current through the meter is simply the voltage divided by its resistance.
If you can't get the answer from this (I got the answer you gave), tell us where you are having trouble.

 

[AnthonyR23]

Thanks!!! I figured it out!! Much appreciated!!!!!