JAMB - Physics (1983 - No. 17)
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
\(\frac{5}{99}\)ohms in series with it
\(\frac{5}{99}\)ohms in parallel with it
\(\frac{99}{5}\)ohms in series with it
\(\frac{99}{5}\)ohms in parallel with it
2 ohms in series with it
Explanation
I = 1 - (10 x 10-3)
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = \(\frac{5}{99}\)\(\Omega\)
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = \(\frac{5}{99}\)\(\Omega\)
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