A moving coil galvanometer is converted into an ammeter, reading upto 0.04 A by connecting a shunt of resistance ' 3r ' across it and then into an ammeter reading upto 0.08 A, when a shunt of resistance ' r ' is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used? (a) 0.02 A (b) 0.04 A (c) 0.08 A (d) 0.01 A

Answer: (a). Shunt rule: Ig G = (I - Ig) S. Ig G = (0.04 - Ig) 3r and Ig G = (0.08 - Ig) r Equate: 0.12 - 3Ig = 0.08 - Ig ⇒ 2Ig = 0.04 Ig = 0.02 A. So the answer is (a).

A moving coil galvanometer is converted into an ammeter, reading upto… | MHT CET 2021 (21 Sep, Shift 2)

Q128MHT CET 2021 · 21 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

A moving coil galvanometer is converted into an ammeter, reading upto $0.04\text{ A}$ by connecting a shunt of resistance ' $3r$ ' across it and then into an ammeter reading upto $0.08\text{ A}$ , when a shunt of resistance ' $r$ ' is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?

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Answer: (a)

Shunt rule: $I_g G = (I - I_g) S$ .
$I_g G = (0.04 - I_g)\,3r$ and $I_g G = (0.08 - I_g)\,r$
Equate: $0.12 - 3I_g = 0.08 - I_g \Rightarrow 2I_g = 0.04$
$I_g = 0.02\text{ A}$ . So the answer is (a).

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