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**1. Homework Statement**

L = 20mH = 20 x 10

^{-3}H

i = 40 mA for t≤0

i = A

_{1}e

^{-10,000t}+ A

_{2}e

^{-40,000t}A for t≥0

The voltage at t=0 is 28 V.

I have to find the equation for the voltage for t>0.

Then I have to find the time when power is zero.

**2. Homework Equations**

v(t) = L*

^{di}/

_{dt}

p(t) = L*i*

^{di}/

_{dt}

**3. The Attempt at a Solution**

I know that current through an inductor has to be continuous, so at t=0, I can say:

A

_{1}+ A

_{2}= 40 x 10

^{-3}A

I then found

^{di}/

_{dt}:

^{di}/

_{dt}= -10,000A

_{1}e

^{-10,000t}- 40,000A

_{2}e

^{-40,000t}

Since v = L

^{di}/

_{dt}:

v = (20 x 10

^{-3}H)*(-10,000A

_{1}e

^{-10,000t}- 40,000A

_{2}e

^{-40,000t})

v = -200A

_{1}e

^{-10,000t}- 800A

_{2}e

^{-40,000t}

At t=0, v=28 V, so:

v(t=0) = -200A

_{1}- 800A

_{2}= 28 V

So I have the following two equations:

A

_{1}+ A

_{2}= 40 x 10

^{-3}A

-200A

_{1}- 800A

_{2}= 28 V

When I solve these using matrices on my calculator, I get:

A

_{1}= 0.1

A

_{2}= -0.06

When I plug those into the equation for v, I get:

v(t) = -20A

_{1}e

^{-10,000t}+ 48A

_{2}e

^{-40,000t}

This answer is one of the multiple choice answers. However, the system tells me this answer is wrong and the correct answer is:

v(t) = 18A

_{1}e

^{-10,000t}+ 10A

_{2}e

^{-40,000t}

I don't understand why my answer is wrong, so any explanation would be appreciated!

Once I understand that, I know that to find the time when power is zero, I just set p(t) = L*i*

^{di}/

_{dt}to zero and solve for t.