Recent content by adrms

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...