A galvanometer having a resistance of 50 Ω, 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× 10-2 cm2 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 ρ=5×10-7 Ω m)

Question

A galvanometer having a resistance of 50 Ω, 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× 10-2 cm2 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 ρ=5×10-7 Ω m) (A) 9 (B) 6 (C) 3 (D) 1.5

Tardigrade Admin · Accepted Answer

S=(Ig G/I-Ig) =(0.05×50/5-0.05) =(2.5/4.95) =(250/495) =(50/99) Ω ∵ S=(ρ l/A) or l=(SA/ρ) ∴ l=(50/99)×(3×10-6/5×10-7) =3.0 m

$9$

$6$

$3$

$1.5$

$S = \frac{I _{g} G}{I - I _{g}}$
$= \frac{0.05 \times 50}{5 - 0.05}$
$= \frac{2.5}{4.95}$
$= \frac{250}{495}$
$= \frac{50}{99} Ω$
$∵ S = \frac{ρl}{A}$ or $l = \frac{S A}{ρ}$
$∴ l = \frac{50}{99} \times \frac{3 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 7}}$
$= 3.0 m$

9

6

3

1.5

S=I−Ig​Ig​G​ =5−0.050.05×50​ =4.952.5​ =495250​ =9950​Ω ∵S=Aρl​ or l=ρSA​ ∴l=9950​×5×10−73×10−6​ =3.0m

$9$

$6$

$3$

$1.5$

$S = \frac{I _{g} G}{I - I _{g}}$
$= \frac{0.05 \times 50}{5 - 0.05}$
$= \frac{2.5}{4.95}$
$= \frac{250}{495}$
$= \frac{50}{99} Ω$
$∵ S = \frac{ρl}{A}$ or $l = \frac{S A}{ρ}$
$∴ l = \frac{50}{99} \times \frac{3 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 7}}$
$= 3.0 m$