Electro Magnetic Induction And Alternating Currents Question 408
Question: Two resistors of 10 $ \Omega $ and 20 $ \Omega $ and an ideal inductor of 10 H are connected to a 2 V battery as . Key K is inserted at time t = 0. The initial (t = 0) and final ( $ t\to \infty $ ) currents through the battery are
Options:
A) $ \frac{1}{15}A,\frac{1}{10}A $
B) $ \frac{1}{10}A,\frac{1}{15}A $
C) $ \frac{2}{15}A,\frac{1}{10}A $
D) $ \frac{1}{15}A,\frac{2}{25}A $
Show Answer
Answer:
Correct Answer: A
Solution:
-
At $ t=0 $ i.e., when the key is just pressed, no current exists inside the inductor.
So $ 10\Omega $ and $ 20\Omega $ resistors are in series and a net resistance of (10+20) = 30 $ \Omega $ exists across the circuit.
Hence, $ I_{l}=\frac{2}{30}=\frac{1}{15}A $
As $ t\to \infty $ , the current in the inductor grows to attain a maximum value. i.e., the entire current passes through the inductor and no current passes through 10 $ \Omega $ resistor.
Hence, $ I_{2}=\frac{2}{20}=\frac{1}{10}A $
BETA
