Question

6. 0/2 points v Previous Answers SERPSET9 32.P.019. Consider the circuit shown in the figure below. (Assume L = 5.25 mh and R2 = 460 2.) L -100% 24.0 V R Rz (a) When the switch is in position a, for what value of R1 will the circuit have a time constant of 15.2 us? 0.458 Your response differs from the correct answer by more than 10%. Double check your calculations. k2 (b) What is the current in the inductor at the instant the switch is thrown to position b? mA Need Heln? Read It

Answer 1

a)

R1 and R2 are in parallel and their parallel combination is given as

Rp = R1 R2/(R1 + R2)

Rp = 460 R1 /(R1 + 460) Eq-1

L = Inductance = 5.25 mH = 0.00525 H

T = Time constant = 15.2 x 10-6 s

Time constant of L-R circuit is given as

T = L/Rp

15.2 x 10-6 = 0.00525/Rp

Rp = 345.4 ohm

Using Eq-1

460 R1 /(R1 + 460) = 345.4

R1 = 1386.42 ohm

b)

Just before throwing the switch, current flowing through the inductor is given using ohm's law as

i = V/Rp (since inductor behaves as short circuit after long time)

i = 24/345.4

i = 0.0695 A