Exam Tomorrow! Can someone help??

[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...

 

[Mikebits]

Since you know P and R for the resistor you calc for voltage. E= sqr(P*R) = 1.336v~
Now that acts as the E source for the meter of 12.6 ohm. So I =E/R so
1.336/12.6 = .106 or 106ma.

 

[AnthonyR23]

Hey, thanks!!! I am still slightly confused with the E= sqr(P*R) = 1.336v~ part of the question. P being the 930mW and for R .... do I just do the 1/12.6 + 1/1.92 = 0.600198413 ..... then 1/.600198413 = 1.666115702, which I believe is the series equivalent resistance for those two resistors in parallel.... So sqr (930 * 1.66115702) ??? Sorry not really understanding how to get to 1.336 v?? Thanks again!!

 

[AnthonyR23]

Hey, I figured it out!!! Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!