# Circuit analysis - laplace transform

So i have a circuit where R1 = 5 W, R2 = 2 Ω, L = 1 H, C = 1/6 F ja E = 2 V. And i need to figure out what is iL when t=0.5s with laplace transform.
Switch opens when t=0
When t<0 i got iL(0)=1A and Uc(0)=0 for initial values.
When t>0 circuit will look like

And now i got for KVL i got

$$E-U_L-U_R-U_C=0$$ $$E-Li_L'-\frac{i_L}{P}-\int\frac{i_L}{C}=0$$ And now in need to do laplace transform. $$E-L(sI_L-i_L(0))-\frac{I_L}{P}- \frac{1}{C}*(\frac{1}{s}*I_L)=0$$ But now i got stuck here. Thanks in advance.

• The capacitor has 2V at t=0, the same voltage as over R2. – user287001 Nov 6 '17 at 11:40
• Just have to comment saying thank you for showing your attempt first. You would not believe how many people ask homework questions here without showing any of their own work – DerStrom8 Nov 6 '17 at 12:00

Assuming R1=5Ω ,

Initially, $V_C\ + V_{R_1} = V_{R_2}$. Since no current flows through C for $t(0^-)$ , $i_{R_2}=1A$. So, $V_{R_2}=V_C=2V$.

For $t>0$,

$$E=L\frac{di}{dt} + i\cdot R_1 + \frac{1}{C}\int i dt + V_C(0)$$

Taking Laplace Transform,

$$\frac{E}{s} = L(sI(s)-I_L(0)) + I(s) \cdot R_1 + \frac{1}{Cs}I(s) + \frac{V_C(0)}{s}$$

Here $I_L(0) = 1A, V_C(0) = 2V$

Rearranging,

$$\frac{E+LsI_L(0)-V_C(0)}{Ls^2 + s R_1 + \frac{1}{C} } = I(s)$$

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit:

The Laplace Inverse of 1/7 is 1/7δ(t), which is a function that takes the value 0 for all $t\neq 0$.

• Now i got that $$I(s)=\frac{s}{7s+6}$$ ,and then for partial fractions i got $$1/7-\frac{6}{49s+42}$$ and now for inverse transform i have $$\frac{-6}{49}*e^{-(6t/7)}+\frac{1}{7}$$, but now when i put that t=0.5 i got I=-0.15 wich is wrong, do you know what went wrong? – J.Doe Nov 6 '17 at 16:03