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>> No.11667976 [View]
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11667976

>>11667732
First realize that its the same voltage over each component, meaning vC=vR=vL=v. Now sum the currents into the top node and set to zero per Kirchhoff's current law. Get [math] i_R+i_C+i_L=0 [/math]. Now recall formulae for current thru resistor and capacitor. You get [math]
(v/R)+cv'+i_L=0 [/math]. Differentiate this and multiply by RL. You get the differential equation: [math] RLCv''+Lv'+Rv=0 [/math]. The transient response can be found by writing the characteristic equation: [math] RLCs^2+Ls+R=0 [/math]. Solve quadratic and simplify a little to get [eqn] s=-\frac{1}{2RC}\pm\sqrt{\frac{1}{4R^2C^2}-\frac{1}{LC}}=-\alpha\pm\sqrt{\alpha^2+\omega_0^2} [/eqn] Take s1 to be the "plus" part and s2 to be the "minus" part. Then the transient voltage is [math] v(t)=K_1\exp{s_1t}+K_2\exp{s_2t} [/math]. But you already now the current through the capacitor is [math] Cv [/math], so [math] Q(t)=CK_1\exp{s_1t}+CK_2\exp{s_2t}=A\exp{s_1t}+B\exp{s_2t}
[/math]. Now look at initial conditions: charge at t=0 is Q0, so A+B=Q0. Current is initially zero, so s1.A+s2.B=0. I'm not gonna go into these weeds for you but you can solve for A and B and get a governing differential equation for charge: [eqn] Q(t)=Ae^{s_1 t}+Be^{s_2 t} [/eqn] The decay constant is [math] \alpha=1/2RC [/math] and the natural frequency is [math] \omega_0=(LC)^{-1/2} [/math]. You get the circuit is overdamped if the determinant of the characteristic equation is greater than zero. So let [eqn] \zeta\equiv\frac{L}{4R^2C} [/eqn] If zeta is greater than 1 its overdamped, less than 1 is underdamped, and unity means critically damped.
>>11667892
AAAAAAAAAAAAAAAAAA
yES.

>> No.11630510 [View]
File: 132 KB, 423x342, K.png [View same] [iqdb] [saucenao] [google]
11630510

>>11630477
Imagine an electron floating in space. There will be an electric field around it but no magnetic field. Or, consider a bar magnet. There will be magnetic flux in places around it, but no electric. That equation clearly doesn't hold.
>but google said it
>pic rel
>>11630482
Then figure out what the directions are. Energy flux only makes sense if you have a direction or oriented surface in mind.

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