Find max. stored energy (in capacitance and inductance)

In summary: Since the voltage across an inductor is proportional to its inductance and the time derivative of the current, we can substitute V=L\frac{di}{dt}=RI to get the expression for power. Integrating this over time gives the expression for energy. In summary, the max. stored energy for a capacitance C is \frac{1}{2}CV_m^2 and for an inductance L is \frac{1}{2}LI^2. In both cases, this comes from the formula for energy stored in the component, with a factor of 1/2 appearing because the power is the time derivative of the energy. Additionally, the R in the formula for the inductance comes from Oh
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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). :redface:
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]
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]


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).
An 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]
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 power
[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?
 
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  • #2
The max. stored energy in an inductor is \frac{1}{2}LI^2. This comes from the formula for the energy stored in an inductor, given by W=\frac{1}{2}LI^2. The factor of 1/2 appears because the power is the time derivative of the energy, and the time derivative of a constant is 0. The R in the formula comes from Ohm's law: V=IR.
 

1. How do you calculate the maximum stored energy in a capacitor?

The maximum stored energy in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage across the capacitor.

2. What factors affect the maximum stored energy in a capacitor?

The maximum stored energy in a capacitor is affected by the capacitance and the voltage. A higher capacitance or voltage will result in a higher maximum stored energy.

3. How do you find the maximum stored energy in an inductor?

The maximum stored energy in an inductor can be calculated using the formula E = 1/2 * L * I^2, where E is the energy in joules, L is the inductance in henries, and I is the current flowing through the inductor.

4. What is the relationship between maximum stored energy and frequency in an inductor?

The maximum stored energy in an inductor is directly proportional to the frequency of the current passing through it. This means that a higher frequency will result in a higher maximum stored energy.

5. How can the maximum stored energy in a capacitor or inductor be increased?

The maximum stored energy in a capacitor or inductor can be increased by increasing the capacitance or inductance, respectively, or by increasing the voltage or current. This can be achieved by using larger components or by using a higher power source.

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