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A galvanometer having a resistance of 9 ohm is shunted by a wire of resistance 2 ohm. If the total current is 1 amp, the part of it passing through the shunt will be

Asked in AIPMT 1998 · Conversion to ammeter and voltmeter

Answer: (2) 0.8 amp

Step-by-step solution

Given: G=9 Ω and S=2 Ω in parallel, total current 1 A.

Idea: in a parallel pair the current divides inversely as the resistances — the easier path takes more.

The share through the shunt is G/(G+S) of the total.

Iₛ=1×9/(9+2)=9/(11).

=0.818 A.

To the figures the options carry, 0.8 A.

The remaining 0.18 A goes through the galvanometer, the higher-resistance branch, as it should.

Why the other options are wrong

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