How does the current divide in parallel inductors?

[jangchen]

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

 

[DaveE]

The current divider rule assumes a steady state solution. This is a transient problem. The current divider rule doesn't properly account for the initial conditions. At t=0 each inductor has 2A, which isn't consistent with the 3:5 ratio.

 

[ehild]

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

The voltage divider rule is valid for resistors, when the voltage is proportional to the current, U=RI. In case of inductors, this is not true, U=LdI/dt instead.

 

[jangchen]

The current divider rule assumes a steady state solution. This is a transient problem. The current divider rule doesn't properly account for the initial conditions. At t=0 each inductor has 2A, which isn't consistent with the 3:5 ratio.

Thank you for your advise! I got what is a steady state right away.

 

[jangchen]

The voltage divider rule is valid for resistors, when the voltage is proportional to the current, U=RI. In case of inductors, this is not true, U=LdI/dt instead.

Oh, I thought current divider was also applied to the inductor. Thank you for your help!