In fig 3165 g 100 v r1 100 ohm r3 300 ohm and l 200 h fi

In fig 31-65, g = 100 v, r_1 = 10.0 ohm, r_3 = 30.0 ohm and l = 2.00 h find the values of l_1 and l_2 impediment after the closing of switch s a long time after the reopening of switch and a long time the reopening.

Solution

(A) Immediately after closing of switch:

no current will flow through the inductor.

i1 = i2 = E / (R1 + R2) = 100 / (10 + 20) = 3.33 A

(B) after long time,

now there will no PD across inductor. (it will behave as simple wire with no resistance)

R3 and R2 are in parallel connection.

1/R\' = 1/R2 + 1/R3 = 1/20 + 1/30

R\' = 12 ohm

now R1 and R\' are in series.

I = V / (R + R\') = 100 / (12+10) = 4.54 A

i1 = I = 4.54 A ..........Ans

i2 = I x R3 / (R2 + R3) = 4.54 x 30 / (20 + 30) = 2.73 A .......Ans

(C) now current will through R2 and R3.

i1 = 0

i2 = 4.54 - 2.73 = 1.81 A

(D) AFTER long time,

i1 = i2 = 0