Exam

JEE Main

Year

2015

Date

4 April

Subject

Physics

Chapter

Alternating Current

JEE Main 2015 · 4 April · Physics · Alternating Current

An inductor ( L = 0.03H ) and a resistor ( R = 0.15kΩ ) are...

An inductor $( L = 0.03\text{H} ) and a resistor$( R = 0.15\text{kΩ} )$are connected in series to a battery of$15\text{V}$E.M.F. 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$t = 1\text{ms}$, the current in the circuit will be : \left(\text{Take}, {e}^{5} \approx 150\right)$

Step-by-step solution

Case I: ${K}_{1}$~is closed for long time,&



for& a long time, the inductor acts as a conducting wire,& $⇒$current in the circuit $= \dfrac{V}{R}= \dfrac{1 5}{1 5 0}\text{⇒}{i}_{0} = \text{0.1 A}$
Case II: ${K}_{1}$~is open and ${K}_{2}$~is closed



Current in the circuit,& $\text{i} = {i}_{0}\text{e}\dfrac{- \text{t}}{τ}\text{;}τ = \dfrac{L}{R}$

After $\text{t} = 1\text{ms} = 1{0}^{- 3}\text{s}$

$\text{⇒}\text{i} = {i}_{0}e^{- \left(\dfrac{1{0}^{- 3} \times 150}{3 \times 1{0}^{- 2}}\right)}= \text{0.1} \text{e} \dfrac{- 1 5}{3}= \text{0.1}\dfrac{1}{\text{e}5} = \dfrac{0.1}{150}\text{A}= \text{0.67} mA$.