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Q.

When a galvanometer having coil resistance 100Ω is converted into ammeter, 90% of the main current is by passed through the shunt, then resistance of the ammeter is

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a

10 Ω

b

5 Ω

c

90 Ω

d

9 Ω

answer is A.

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Detailed Solution

Rsh×0.9 I=100×0.1 I⇒ Rsh=100/9 Ω∴ Resistance of ammeter=100×1009(100+1009)Ω=10Ω
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When a galvanometer having coil resistance 100Ω is converted into ammeter, 90% of the main current is by passed through the shunt, then resistance of the ammeter is