Electro Magnetic Induction And Alternating Currents Question 323
Question: An inductor of 2 henry and a resistance of 10 ohms are connected in series with a battery of 5 volts. The initial rate of change of current is [MP PMT 2001]
Options:
A) 0.5 amp/sec
B) 2.0 amp/sec
C) 2.5 amp/sec
D) 0.25 amp/sec
Show Answer
Answer:
Correct Answer: C
Solution:
$ i=i_{0}( 1-{{e}^{\frac{-Rt}{L}}} )\Rightarrow \frac{di}{dt}=\frac{d}{dt}i_{0}-\frac{d}{dt}i_{0}{{e}^{\frac{-Rt}{L}}} $
$ \Rightarrow \frac{di}{dt}=0-i_{0}( -\frac{R}{L} )\ {{e}^{-\frac{Rt}{L}}}=\frac{i_{0}R}{L}{{e}^{-\frac{Rt}{L}}} $
Initially, $ t=0\Rightarrow \frac{di}{dt}=\frac{i_{0}\times R}{L}=\frac{E}{L}=\frac{5}{2}=2.5\ amp/sec. $
