A milliameter with full scale deflection of 10mA has an internal resistance of 5 ohms. It would be converted to an ammeter with a full scale deflection of 1A by connecting a resistance of
A. \(\frac{5}{99}\)ohms in series with it B. \(\frac{5}{99}\)ohms in parallel with it C. \(\frac{99}{5}\)ohms in series with it D. \(\frac{99}{5}\)ohms in parallel with it E. 2 ohms in series with it
Correct Answer: B
Explanation
I = 1 - (10 x 10-3) = 0.99A IR = 10 x 10 x 10-3 x 5 R = \(\frac{5}{99}\)\(\Omega\)
