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The two parallel inductors are connected across the terminals of a black box at t = 0. The resulting voltage v(t) for t ≥ 0 is known to be 40e-1.25t V. a. Replace the original inductors with an equivalent inductor and find i(t) for t ≥ 0. b. Find i1(t) for t ≥ 0. c. Find i2(t) for t ≥ 0. d. How much energy is delivered to the black box in the time interval 0 ≤ t ≤ [infinity]? e. How much energy was initially stored in the parallel inductors? f. How much energy is trapped in the ideal inductors? g. Do your solutions for i1 and i2 agree with the answer obtained in (f)?

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Answer:

[tex]V(t) = 12e^{-t} V, for t > 0\\i_{1} = 2A\\i_{2} = 4A[/tex]

a) [tex]i(t) = 6e^{-t}[/tex]

b) [tex]4e^{-t}-2A[/tex]

c) [tex]i_{2}(t) = 2e^{-t} + 2A[/tex]

d) 36 joules

e) 48 joules

f)18 joules

g) Yes, they are equal

Explanation:

a) - e) integrate from the circuit diagram

f) Trapped energy by the indicators:

L₁ = 6 joules

L₂ = 12 joules

g) The energy delivered to the Blackbox  = Indicator Initial Energy - Inductor Trapped Energy

This translates to 36 = 54 - 18

                             36  = 36

Therefore, as the left hand side is equal to the right hand side, they are equal.