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3. The attempt at a solution

Just was hoping someone here would be able to help me with the circuit above. I wanted to ensure that I analyzed it properly.

I applied Ohm's Law to determine the initial current through the inductor. i(initial)=-16V/2.2kΩ=-7.27 mA. Negative sign indicating up through the inductor.

Next, I applied Thevenin's Theorm to reduce the circuit to the left of the inductor. The right sight behaves as an open circuit due to the open switch.

I get Rth = 8.2+2.2 k-Ohm=10.4 kΩ

Additionally, I determine the Thevenin equivalent voltage is:

Eth=(4mA)(8.2)=32.8V

I can then determine the equation that describes the current, i, in the system.

i = i(final)+[i(initial)-i(final)]e^(-t/τ)

where i(final) is determined by assuming steady state conditions.

i(final)=32.8V/10.4kΩ=3.1538mA

thus,

i = 3.14mA+[-7.27-3.15]e^(-t/τ) and τ= 500/10.4=48.08 s

and

i=3.15-(10.4)e^(-t/48.08)

I found the voltage to be a bit more tricky and my result seems very high.... here is what i did:

I noted the inital polarity of the inductor to be positive polarity at the ground, and negative at the top. I then applied KVL around the loop to determine the voltage.

Vm = -32.8V - (7.27mA)(10.4kΩ) = -108.4V.

Thus, V=-108e^(-t/48.08)

Thanks a bunch in advance!

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# Circuit Analysis Involving Inductance

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