A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a shunt of resistance 4r across it. and into an ammeter reading up to 0.06 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?

Question

A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a shunt of resistance 4r across it. and into an ammeter reading up to 0.06 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) 0.01 A (B) 0.02 A (C) 0.03 A (D) 0.04 A

Tardigrade Admin · Accepted Answer

For an ammeter (ig/i)=(S/G+S) ⇒ ig G=(i-ig) S ∴ ig G=(0.03-ig) 4 r ldots (i) and ig G=(0.06-ig) r ldots (ii) From Eqs. (i) and (ii) 0.12-4 ig=0.06-ig ⇒ ig=0.02 A

0.01 A

0.02 A

0.03 A

0.04 A

For an ammeter iig​​=G+SS​ ⇒ig​G=(i−ig​)S ∴ig​G=(0.03−ig​)4r… (i) and ig​G=(0.06−ig​)r… (ii) From Eqs. (i) and (ii) 0.12−4ig​=0.06−ig​ ⇒ig​=0.02A

For an ammeter $\frac{i _{g}}{i} = \frac{S}{G + S}$
$\Rightarrow i_{g} G = (i - i_{g}) S$
$∴ i_{g} G = (0.03 - i_{g}) 4 r \dots$ (i)
and $i_{g} G = (0.06 - i_{g}) r \dots$ (ii)
From Eqs. (i) and (ii)
$0.12 - 4 i_{g} = 0.06 - i_{g}$
$\Rightarrow i_{g} = 0.02 A$