Current Electricity Charging Discharging Of Capacitors Question 298
Question: A galvanometer, having a resistance of 50 W 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× 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 = 5 × $ {{10}^{-7}} $ Wm)
[EAMCET 2003]
Options:
A) 9
B) 6
C) 3
D) 1.5
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{i}{i _{g}}=1+\frac{G}{S}\Rightarrow \frac{5}{0.05}=1+\frac{50}{S} $
Therefore $ S=\frac{50}{99}=\frac{\rho \times l}{A} $
Therefore $ l=\frac{50}{99}\times \frac{2.97\times {{10}^{-2}}\times {{10}^{-4}}}{5\times {{10}^{-7}}}=3m $ .
BETA
