# How to know the values of $R$ and $L$?

Yo there! I have a RL circuit in the arrange showed in the image. Ok, the trick is how can determine the values of the components based on the graphics of a step and impulse response, I detail the process and the graps in the images; basically what is done is to determine the model, calculate the unit and step response, got a 3 equations systems, but at the end, it is only a quotient, but aside that I don't see where can it get the values of R and L. Thanks in advance

The graphs are double-checked, it seems odd but are ok. 1st , it gets the math model of the system

$$i(t)\frac{R_{1}}{L}=\frac{dI_{L}}{dt}+(\frac{R_{2}+R_{1}}{L})I_{L}}}}$$

Calculating unit step and impulse response applying the input $i(t)=10_{u}(t)[A]$

$$I_{L}(t)=\frac{10R_{1}}{R_{1}+R_{2}}(1)-\frac{10R_{1}}{R_{1}+R_{2}}e\frac{-R_{1}+R_{2}}{L}t}}}$$ and $$h(t)=\frac{10R_{1}}{L}e\frac{-R_{1}+R_{2}}{L}t}}$$ In the step graph it says $I(t)$ but there is trouble trying to fit the data, perhaps it should say $I_{L}(t)$, taking the graph this way: From the step graph $$\frac{10R_{1}}{R_{1}+R_{2}}=5...[1a]}$$ and $$\frac{10R_{1}}{R_{1}+R_{2}}(1-e^{-\frac{R_{1}+R_{2}}{L}(5\mu s)}})=4.325...[2a]}$$ 1a in 2a $$4.325=5(1-e^{-\frac{R_{1}+R_{2}}{L}(5\mu s)}})...[3a]$$ from the impulse $$68000=\frac{5R_{1}}{L}e^{\frac{-R_{1}+R_{2}}{L}(2.5\mu s)}}...[4a]}$$ from [3a] $$\frac{4.325}{5}=(1-e^{-\frac{R_{1}+R_{2}}{L}(5\mu s)}})...[5a]}$$ Grouping terms and aplying the natural log $$ln(e^{-\frac{R_{1}+R_{2}}{L}(5\mu s)}})=ln(1-\frac{4.325}{5})...[5b]}$$

$$\frac{R_{1}+R_{2}}{L}=\frac{-2.0024}{-5\mu}=400,490...[5c]}}$$ From 1a $$10R_{1}=5(R_{1}+R_{2})...[6a]5R_{1}=5R_{2}...[6b]R_{1}=R_{2}...[6c]$$ from 6c in 5c $$\frac{2R}{L}=400,490}$$ $$\frac{R}{L}=200,245...[7a]}$$

This one is where it gets stalled because I don't see where to tackle the values,

• Yo yourself : ), I added in the image because users normally like to read without going off site. Personally, I think this is an ElectronicsSE kind of question, but the best of luck with it. – user108787 Oct 3 '16 at 16:39
• Thanks CountTo10, It was my idea to add the picture, but the editing form says I cannot insert directly images. Thanks a lot again. – riccs_0x Oct 3 '16 at 17:35