A 9.0-V battery is connected through a switch to two identical resistors and an

Question

A 9.0-V battery is connected through a switch to two identical resistors and an ideal inductor as shown below. Each of the resistors has a resistance of 100 Ohm and the inductor has an inductance of 3.0 H. The switch is initially open. Immediately the switch is closed (at t = 0), find: l_1, I_2 and l_3 The potential difference across R_1 and R_2 The potential difference across the inductor L Now the switch has been closed for a very long time, now calculate: I_1, I_2 and l_3 The rate of current change across the resistor R_2 The potential difference across L

Explanation / Answer

first you have to know the behaviour of inductors in DC circuit

it behave like broken wire just after circuit closed.

& it behave like conducting wire after long time .

case 1st immediately ( it behave   like broken)

so I3 = 0

i1 = i2 = V/R = 9/ 100 = 0.09 A

Vr1 = 9 V

Vr2 = 0V

potnetial across L = 9 V

case 2 very long time ( it behave  like conducting wire)

i3 = 9/ 100 = 0.09

I2 = 9/ 100 = 0.09

i1 = i2+i3 = 0.18

i =V/R ( 1- e^(-RT/ L ) )

di/dt = ? ...........answer

total R = 50 ohm

potential across L = 0 V

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