# 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. 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 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? Nov 6 '17 at 16:03