A galvanometer having a resistance of 8 ohm is shunted by a wire of resistance 2 ohm. If the total current is 1 A, the part of it passing through the shunt will be (a) 0.25 A (b) 0.8 A (c) 0.2 A (d) 0.5 A

Answer: (b). Current divides inversely as resistance: IS = I(G)/(G + S). IS = 1 × (8)/(8 + 2) = 0.8 A So the answer is (b) 0.8 A.

A galvanometer having a resistance of 8 ohm is shunted by a wire of… | NEET 1998 (AIPMT)

Q205NEET 1998 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments

A galvanometer having a resistance of $8$ ohm is shunted by a wire of resistance $2$ ohm. If the total current is $1\text{ A}$ , the part of it passing through the shunt will be

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Answer: (b)

Current divides inversely as resistance: $I_S = I\frac{G}{G + S}$ .
$I_S = 1 \times \frac{8}{8 + 2} = 0.8\text{ A}$
So the answer is (b) $0.8\text{ A}$ .

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