First the hybrid model, I assume the capacitor works as a short circuit regarding the altern current:
$$A_V=\frac{v_o}{v_i}$$
$$v_i=-v_{GS}$$
$$v_o=-i_LR_L$$
$$i_L=g_mv_{GS}+\frac{v_{DS}}{r_d}$$
Now I use the Kirchhoff's law to get $v_{DS}$. I consider this close loop:
$$-v_i-v_{DS}+v_o=0$$...
I thought that I had understood it well, please let me explain in detail my procedure to solve the problem.
First I will answer the practice example 4 to be sure that I understand where RMS value of a signal comes from.
Let me see, I can derive it seeing the other examples.
x_{\text{peak}} =...
Oh, I see...
So the equation of the voltage signal is:
$$V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$$
$$V = v(0.2 \cos (4t)) (2 \pi r) + 0.8 \sin (4t) (\pi r ^2)$$
Replacing the radius r=0.5m
$$V = 0.2 \pi v \cos (4t) + 0.2 \pi \sin (4t)$$
Then I can "merge" the sinusoidal terms using...
First of all, thank you for welcoming me to the forum 😁
I was not aware of that fact. Thank you for clarifying it.
I'm sorry, I think I may have messed it up while taking the derivative and replacing the variables.
Please let me start with this equation:
v = \frac{V + (\pi r ^2)...
$$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \int _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$
From the statement I know that: B⊥v, (B x v) // dl and B // ds.
$$V = \int vBdl - \oint _S \frac{dB}{dt} ds$$
v is the speed with which all the segments dl are aproximating to the...