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
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Correct Answer: Option B
Explanation:
I = 1 - (10 x 10-3)
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = \(\frac{5}{99}\)\(\Omega\)
I = 1 - (10 x 10-3)
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = \(\frac{5}{99}\)\(\Omega\)