# I am trying to find equations for i1(t) and i2(t) using laplace transform for this step input

The place that I am confused is that for $$\i_1(t)\$$ im trying to use the loop method to develop the first equation which is

$$v(t)-L_1\frac{di}{dt}-(R_1+R_2)i_1(t)=0$$

and then use $$\i_2(t)\$$ for the second loop

$$R_3i_2(t)+L_2\frac{di}{dt}+R_4i_2(t)=0$$

Is this the correct approach and if yes are the equations correct for $$\i_1\$$ and $$\i_2\$$ ?

Thank you for any guidance.

• Replace the inductor with its Laplace equivalent, sL, and treat like a ‘normal’ circuit analysis problem, with the step input, V/s
– Chu
Apr 1 at 21:50

It might be better to draw the currents as circular within the mesh, because it looks like you forgot that R1 and R2 use both $$\i_1(t)\$$ and $$\i_2(t)\$$:

Now you can see that the equation for the first loop is:

$$v_1(t)=L_1\dfrac{\text{d}i_1(t)}{\text{d}t}+(R_1+R_2)\biggr(i_1(t)-i_2(t)\biggr)$$

You should be able to continue with the second equation now.

• Thank you this was very helpful Apr 2 at 16:21
• @ChMaryam You're welcome. I'd still use the Laplace analysis first, then inverse Laplace (as mentioned by the other answers and comments) because it's easier to solve. The only reason why I kept it in the time domain was because you seemed to want to continue like this. Apr 2 at 16:37

First, since we are only looking for the two currents we can simplify by lumping some of the resistances together. The Laplace transformed circuit is below.

Solve with any circuit analysis techniques you like and then inverse Laplace transform,$$\\mathscr{L}^{-1}\$$, to find the time-domain result.

• I knew that in laplace the components can be added easily just like resistors but was not sure if that was the approach to take but this does make it easier for me, thank you! Apr 2 at 16:23

Well, I'll solve $$\\text{I}_1\left(t\right)\$$:

$$\text{I}_1\left(t\right)=\int_0^t\text{V}_\text{i}\left(\tau\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\text{sL}_1+\frac{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4+\text{sL}_2\right)}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4+\text{sL}_2}\right)^{-1}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag1$$

• I'll compare my answer to this and find out the where i went wrong thank you Apr 2 at 16:24