Question

A galvanometer having a resistance of 50$\Omega$ gives a full scale deflection for a current of 0.05 A. The length in meter of a resistance wire of area of cross section $2.97\times {10}^{-3}c{m}^{-2}$ that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is
[specific resistance of the wire $=5\times {10}^{-7}\Omega m$]

• A. 9
• B. 6
• C. 3
• D. 1.5

Answer 1

Answer: (C)

3

The shunt resistance used is given by:

S = $\frac{I_{g}G}{I-I_g}$

=$\frac{0.05\times 50}{5-0.05}=\frac{2.5}{4.95}=\frac{250}{495}=\frac{50}{99}\Omega$

We know that,

$\because$ S=$\frac{\rho l}{A}$ or $l=\frac{SA}{\rho }$

$l=\frac{50}{99}\times \frac{3\times {10}^{-6}}{5\times {10}^{-7}}=3.0m$