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## Homework Statement:

- In the circuit of Fig. 63.82 io(0)=2mA. Determine io(t) and Vo(t) for t>0

## Relevant Equations:

- $$V_O = L*\frac{di}{dt}$$

I apologize using English fluently because I am not an Enlgish speaker.

When I tried to solve this problem, I used current divider rule.

So, $$i_o(t) = \frac{3}{3+5}*4e^{-2t} = 1.5*e^{-2t} A$$

However, This was wrong.

The answer is $$ 1.5*e^{-2t} + 0.5 A$$

If I use $$V_O = L*\frac{di}{dt}$$ , I can get right answer.

I wonder why there is a difference

between using current divider rule and using $$V_O = L*\frac{di}{dt}$$.