Question

Two resistors of `10 Omega` and `20 Omega` and an ideal inductor of `10 H` are connected to a `2 V` battery as shows in Fig. key `K` is inserted at time `t = 0`. The initial `(t = 0)` and final `(t rarr oo)` currents through the battery are

A. `(1)/(15) A`,`(1)/(10) A`
B. `(1)/(10) A`,`(1)/(15) A`
C. `(2)/(15) A`,`(1)/(10) A`
D. `(1)/(15) A`,`(2)/(25) A`

Answer 1

Correct Answer - A
At `t = 0`,i.e., when the key is just pressed, no current exiests inside the inductor. So `10 Omega` and `20 Omega` resistance are in series and a net resistance of `(10 + 20) = 30 Omega` exists across the circuit.
Hence, `I_(1) = (2)/(30) = (1)/(15) A`
As `t rarr oo`, the current in the inductor grows to attain a mximum value, i.e., the entrie current passes through the inductor and no current passes through `10 Omega` resistor.
Hence, `I_(2) = (2)/(20) = (1)/(10) A`