I need help with this problem. A 9.0-V battery is connected through a switch to
ID: 1516782 • Letter: I
Question
I need help with this problem.
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 Qand 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 I_3 The potential difference across R_1and R_2 The potential difference across the inductor L Now the switch has been closed for a very long time, now calculate: l_1, l_2and l_3 The rate of current change across the resistor R_2 The potential difference across LExplanation / Answer
irst 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