Okay. I think I got it.
Now, using KCL at the node above the dependent current source I have:
0 = 4 - 3_i_1 - i_1
4 = 4_i_1
\ldots i_1 = 1 amp
V_{oc} = 5 V \ldots V_T = 5 V
I_{sc} = 4 A
R_{th} = \frac{V_{oc}}{I_{sc}}
R_{th} = \frac{5}{4} \Omega
So, a voltage source of 5V...
Homework Statement
Basically asking me to find the Thevenin/Norton's equivalent across a-b.
Then figure out the maximum power transferred
Then figure out max power delivered
Homework Equations
Maximum power is transferred when [SIZE="5"]R_{l} = R_{th} .
Max power delivered is...
I just need a hint or something to see where I start. I'm at a loss for a beginning.
Consider the non-homogenous equation
y'' + xy' + y = x^2 +2x +1
Find the power series solution about x=0 of the equation and express your answer in the form:
y=a_0 y_1 + a_1 y_2 + y_p
where a_0 and...
I have looked all over the internet and in my book and I cannot find the above mentioned Laplace Transform formula anywhere. Did you mean (n!) in the numerator vice the (n+1) ? In that case, that would make my answer 3t^2
Yes this is where I was lost because I didn't have that in the table my professor gave me. His said:
L\{t^n\}=\frac{\Gamma(n!)}{s^{n+1}}
I am lost because I only see this problem as:
L\{t^n\}=\frac{\Gamma(2+1)}{s^{2+1}}
= t^2... i know this isn't right.
wait... is this what I...
Okay, I know this is alot... but I am stuck, so here goes...
Use the method of Laplace transform to solve the initial value problem
y''+3ty'-6y=0, y(0) = 1, y'(0) = 0
L\{y'' + 3ty' - 6y\} = L\{0\}
s^{2}Y(s) - sy(0) - y'(0) + 3L\{ty'\} - 6Y(s) = 0
s^{2}Y(s) - s(1) - 0 -...
Okay, I know this is alot... but I am stuck, so here goes...
Use the method of Laplace transform to solve the initial value problem
y''+3ty'-6y=0, y(0) = 1, y'(0) = 0
L\{y'' + 3ty' - 6y\} = L\{0\}
s^{2}Y(s) - sy(0) - y'(0) + 3L\{ty'\} - 6Y(s) = 0
s^{2}Y(s) - s(1) - 0 -...