Electro Magnetic Induction And Alternating Currents Question 348
Question: A 50 volt potential difference is suddenly applied to a coil with $ L=5\times {{10}^{-3}} $ henry and $ R=180,ohm $ . The rate of increase of current after 0.001 second is [MP PET 1994]
Options:
A) 27.3 amp/sec
B) 27.8 amp/sec
C) 2.73 amp/sec
D) None of the above
Show Answer
Answer:
Correct Answer: D
Solution:
The rate of increase of current
$ =\frac{di}{dt}=\frac{d}{dt}i_{0}( 1-{{e}^{-Rt/L}} )=\frac{d}{dt}i_{0}-\frac{d}{dt}i_{0}{{e}^{-Rt/L}} $
$ =0-i_{0}{{e}^{-Rt/L}}.\frac{d}{dt}( -\frac{Rt}{L} )=i_{0}\frac{R}{L}{{e}^{-Rt/L}} $
$ =\frac{50}{180}\times \frac{180}{5\times {{10}^{-3}}}\times {{e}^{-(180\times 0.001)/(5\times {{10}^{-3}})}} $ $ =10^{4}\times {{e}^{-36}}A/\sec $
