Asked by abdul rahman

Consider the LR circuit. We have R1=19 Ohms, R2=23 Ohms, R3=23 Ohms, V=14 Volts, L=0.09 H.

L and R2 are connected in series, Combination of L and R2 in series is connected with R3 in parallel. This whole combination is connected in series with R1 and voltage source and switch.

At t=0 the switch is closed.
(a)Immediately after the switch is closed, what are currents I1, I2, I3 ?

(b)What are I1, I2, I3 after the switch has been closed for a long time?

The switch is now opened again.
(c) What are the I1, I2, I3 currents after the switch is reopened?
Answered by Henry
a. I1=I3 = E/(R1+R3)=14/(19+23)=0.333A.
I2 = 0

b. The inductor is fully charged and
acts like a short circuit(Rc = 0).

Rt=R1 + (R2*R3)/(R2+R3)=Tot. Resistance.
Rt = 19 + (23*23)/(23+23) = 30.5 Ohms.

I1 = E/Rt = 14/30.5 = 0.46A

I2 = I3 = I1/2 = 0.46/2 = 0.23A.

c. I1 = 0

I2 = I3 = 0.23A(t = 0).
Answered by abdul rahman
Thnx :)
Answered by dd
C) i3 please, i3 isnt same to i2
Answered by dd
i got it, is the negative :)
Answered by Anonymous
I am not getting it right with other values.
There are no AI answers yet. The ability to request AI answers is coming soon!