1
\$\begingroup\$

enter image description here

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 enter image description here

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.

\$\endgroup\$
2
  • \$\begingroup\$ The capacitor has 2V at t=0, the same voltage as over R2. \$\endgroup\$ – user287001 Nov 6 '17 at 11:40
  • 2
    \$\begingroup\$ 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 \$\endgroup\$ – DerStrom8 Nov 6 '17 at 12:00
0
\$\begingroup\$

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\$.

\$\endgroup\$
1
  • \$\begingroup\$ 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? \$\endgroup\$ – J.Doe Nov 6 '17 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.