Electro Magnetic Induction And Alternating Currents Question 204
Question: An inductor (L=0.03 H) and a resistor $ (R=0.15k\Omega ) $ are connected in series to a battery of 15V emf in a circuit shown below. The key $ K_{1} $ has been kept closed for a long time. Then at t= 0, $ K_{1} $ is opened and key $ K_{2} $ , is closed simultaneously. At f = 1 ms, the current in the circuit will be : $ (e^{5}\cong 150) $
Options:
A) 6.7 Ma
B) 0.67 mA
C) 100 mA
D) 67 mA
Show Answer
Answer:
Correct Answer: B
Solution:
-
$ I(0)=\frac{15\times 100}{0.15\times 10^{3}}=0.1A $ $ I(\infty )=0 $
$ I(t)={{[I(0)-I(\infty )]}_{{{e}^{^{\frac{-t}{L/R}+i(\infty )}}}}} $
$ I(t)=0.1{{e}^{^{\frac{-t}{L/R}+i(\infty )}}}=0.1{{e}^{\frac{R}{L}}} $
$ I(t)=0.1e\frac{0.15\times 1000}{0.03}=067mA $
