Consider the circuit shown in the figure below where L = 4.80 mH and R2 =...

Consider the circuit shown in the figure below where L = 4.80 mH and R2 = 420 ?.

Consider the circuit shown in the figure below whereL=4.80mHandR₂=420?.

(a) When the switch is in position a, for what value of R1 will the circuit have a time constant of 15.1 s k?

(b) What is the current in the inductor at the instant the switch is thrown to position b?

Solutions

Expert Solution

a)time constant

τ =L/R

15.1*10^-6 =4.8*10^-3/R

R=317.88Ω

since R₁ andR₂ are in parallel

R=R₁R₂/(R₁+R₂₎

317.88 =R₁*420/(R₁+420)

317.88R₁+133509.6=420R₁

102.12R₁ =133509.6

R₁ =1307.4Ω≈1.307KΩ

b)I=V/R =24/317.88 =7.55mA

genius_generous answered 2 years ago

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