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
Verified Correct Answer: B.
Physical Principles & Numerical Solution
I = 1 - (10 x 10-3)
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = \(\frac{5}{99}\)\(\Omega\)
Source: Myschool Past Questions Archive (Record #28842)