A sensitive galvanometer has resistance 150 W and requires 0.5 mA of current to

Question

A sensitive galvanometer has resistance 150 W and requires 0.5 mA of current to produce full scale deflection. What resistance would be required to construct a voltmeter reading 5.0 V full scale from this galvanometer? I used the equations I = P/V and I=V/R to get: I = v^2/P = (5^2)/(150) = .16667 R=V/I = 5/.166667 ~ 30 ohms The answer should be 10 MW... This in itself confused me because I thought resistance was supposed to be in ohms so then I tried calculating it a different way and kept the resistance in watts: R = V/I = 5/.0005 = 10,000 W This was closer but also wrong because the answer is supposed to be 10 MW or 10,000,000 W! Please let me know if you can explain what I'm doing wrong.. thanks!

Explanation / Answer

v=Ig(G+R)

R= (v/Ig) -G

= 5/.00005 - 150

= 9850 ohm

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