Electrical engineering circuit analysis

[circuit_boy]

Homework Statement

Question 1: http://img690.imageshack.us/i/79806671.jpg/

Question 2: http://img833.imageshack.us/i/48920688.jpg/

Homework Equations

V=IR, P=I^2R

The Attempt at a Solution

Answer to Question 1: http://img831.imageshack.us/i/imgmp.jpg/

Answers to Question 2 Part 1 :http://img545.imageshack.us/i/img0001e.jpg/

Answers to Question 2 Part 2 :http://img94.imageshack.us/i/img0002te.jpg/

Here are my answers for Question 2, if the solution is too long for you to bother with =D
a) Vth = 13V, Isc = 13A, Rth = 1Ohm
b) Vx = 12V
c) Vx = 12V
d) 8 watts, 2 watts, and 12 watts
e) Independent sources 16V and 10V absorb power, dependent source 1.2Vx delivers power

Feel free to give me any inputs! Your help is really appreciated!

 

[lewando]

At t=3, Q=5. How do you get a negative current?

 

[circuit_boy]

Since Q/T = I

We can say that taking the derivative of Q with respect to T will yield I, which is actually the gradient of Q and T, or of the graph.

From t=2 till t=4, gradient is negative, hence yielding a negative value. Are there anymore errors that you have spotted in the solution?

 

[collinsmark]

Homework Statement

Question 1: http://img690.imageshack.us/i/79806671.jpg/

[...snip...]

The Attempt at a Solution

Answer to Question 1: http://img831.imageshack.us/i/imgmp.jpg/

In the future, please post only one problem per thread (different questions related to the same problem are okay, but just don't post completely different problems). It gets really confusing if people are trying to help you with different problems all in the same thread.

So for "Question 1" (actually, Problem# 1), your solutions to parts A and B are correct.

Part C, the rms current is incorrect.
For RMS,
(1) First take the square of the function
(2) Find the "mean" of that squared function, i.e. the average of the squared function. This generally done by integrating the squared function (i.e. finding the area under the curve of the squared function) and then dividing by the time.
(3) Take the square root of whole result.

In your particular problem, you didn't find the mean correctly. (The mean of a and b is not a/2. The mean is (a + b)/2).

More generally, rms for a continuous function is:

f_{rms} = \sqrt{\frac{1}{T_2-T_1}\int_{T_1} ^{T_2}[f(t)]^2 dt}

 

[circuit_boy]

I see, the answer would be 5 then? Thank you for your input, try to have a look at question 2 too, thank you!