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Find "max. stored energy" (in capacitance and inductance)
1.) I'll start with capacitance since I get closer to solution (than in inductance problem).
Now combine the two into the power formula ...
[tex]p(t)=i(t) v(t)=-\frac{V_m^2}{R}e^{-\frac{2t}{RC}}[/tex]
... and integrate it to [tex]\infty[/tex] to get "max. stored energy": [tex]E=-\frac{V_m^2}{R}\int_0^\infty e^{-\frac{2t}{RC}}dt=-\frac{V_m^2}{R}(-\frac{RC}{2})e^{-\frac{2t}{RC}}|_0^\infty=-\frac{1}{2}CV_m^2[/tex]
Where did that extra minus come from?
2.) The second problem, this time with inductance. Here, I'm not sure what exactly is "max. stored energy" as inductor "stores" energy in magnetic field (whatever that means to a novice like me).
[tex]p(t)=u(t)i(t) = (L\frac{di(t)}{dt}) i(t) = RI^2(e^{-\frac{Rt}{L}}-e^{-2\frac{Rt}{L}})[/tex]
So, I derived [tex]\frac{dp(t)}{dt}=\frac{R^2I^2}{L}(2e^{-2\frac{Rt}{L}}-e^{-\frac{Rt}{L}})=0 \Longleftrightarrow 2e^{-2\frac{Rt}{L}}=e^{-\frac{Rt}{L}} \rightarrow \frac{Rt}{L}=\ln(2)[/tex]
Inserting back into [tex]p(t)[/tex]:
[tex]p(t=\frac{L}{R}ln(2))=RI^2(1-e^{-\ln(2)})e^{-\ln(2)}=\frac{1}{4}RI^2 \neq \frac{1}{2}LI^2[/tex]
Why do I get 1/4 instead of 1/2 ... and why R instead of L?
1.) I'll start with capacitance since I get closer to solution (than in inductance problem).
First get the voltage [tex]v(t)=\frac{1}{C}\int i(t)dt=\frac{1}{C}\int \frac{V_m}{R}e^{-\frac{t}{RC}}=-V_m e^{-\frac{t}{RC}}[/tex]A capacitance [tex]C[/tex] has a current [tex]i=\frac{V_m}{R}e^{-\frac{t}{RC}}[/tex]. Find the max. stored energy, assuming zero initial charge.
Ans. [tex]\frac{1}{2}CV_m^2[/tex]
Now combine the two into the power formula ...
[tex]p(t)=i(t) v(t)=-\frac{V_m^2}{R}e^{-\frac{2t}{RC}}[/tex]
... and integrate it to [tex]\infty[/tex] to get "max. stored energy": [tex]E=-\frac{V_m^2}{R}\int_0^\infty e^{-\frac{2t}{RC}}dt=-\frac{V_m^2}{R}(-\frac{RC}{2})e^{-\frac{2t}{RC}}|_0^\infty=-\frac{1}{2}CV_m^2[/tex]
Where did that extra minus come from?
2.) The second problem, this time with inductance. Here, I'm not sure what exactly is "max. stored energy" as inductor "stores" energy in magnetic field (whatever that means to a novice like me).
As I said, I'm not sure about my intuition of how inductor "stores" energy ... I imagined that the "max. stored energy" is the instantaneous peak in powerAn inductance [tex]H[/tex] has a current [tex]i=I(1-e^{-\frac{Rt}{L}})[/tex]. Find the max. stored energy.
Ans. [tex]\frac{1}{2}LI^2[/tex]
[tex]p(t)=u(t)i(t) = (L\frac{di(t)}{dt}) i(t) = RI^2(e^{-\frac{Rt}{L}}-e^{-2\frac{Rt}{L}})[/tex]
So, I derived [tex]\frac{dp(t)}{dt}=\frac{R^2I^2}{L}(2e^{-2\frac{Rt}{L}}-e^{-\frac{Rt}{L}})=0 \Longleftrightarrow 2e^{-2\frac{Rt}{L}}=e^{-\frac{Rt}{L}} \rightarrow \frac{Rt}{L}=\ln(2)[/tex]
Inserting back into [tex]p(t)[/tex]:
[tex]p(t=\frac{L}{R}ln(2))=RI^2(1-e^{-\ln(2)})e^{-\ln(2)}=\frac{1}{4}RI^2 \neq \frac{1}{2}LI^2[/tex]
Why do I get 1/4 instead of 1/2 ... and why R instead of L?