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| 1. |
A galvanometer has a resistance of 100 Omega. A current of 2 xx 10^(-3) A can pass through the galvanometer. How can it be converted into (a) Ammeter of range 20 A and (b) oltmeter of range 20 V? |
| Answer» Solution :`(100)/(999) OMEGA (B) 9900 Omega` | |