Question

$A$ galvanometer having a resistance of $50 \, \Omega$, gives a full scale deflection for a current of $0.05 \,A$. The length (in metres) of a resistance wire of area of cross section $3\times 10^{-2}\, cm^{2}$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \,A$ current is (Specific resistance of the wire $\rho=5\times10^{-7}\,\Omega\,m)$

• A. $9$
• B. $6$
• C. $3$
• D. $1.5$

Answer 1

Answer: (C)

$3$

Answer 2

$S=\frac{I_{g}\,G}{I-I_{g}}$ $=\frac{0.05\times50}{5-0.05}$ $=\frac{2.5}{4.95}$ $=\frac{250}{495}$ $=\frac{50}{99}\, \Omega$ $\because S=\frac{\rho l}{A}$ or $l=\frac{SA}{\rho}$ $\therefore l=\frac{50}{99}\times\frac{3\times10^{-6}}{5\times10^{-7}}$ $=3.0\,m$