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This question formed while trying to size a NTC thermistor to limit the inrush current to a capacitor. Assume the schematic below (with a fixed resistance):

schematic

simulate this circuit – Schematic created using CircuitLab

The total energy \$ E \$ provided by the inrush current to the capacitor is given by:

$$ E = \frac{1}{2} C V_{in}^2 $$

where:
\$ C \$ = the downstream capacitance, in Farads
\$ V_{in} \$ = the input voltage, in Volts

Many sources (http://powerelectronics.com/community/how-do-you-choose-right-type-ntc-thermistor-limit-inrush-current-capacitive-applications, http://www.ametherm.com/inrush-current/) use this equation to work out how much energy the NTC thermistor will absorb at turn on (when SW1 is closed).

This is where I am confused. I thought this equation tells you how much energy flows through the thermistor and is delivered to the capacitor. The NTC thermistor will absorb some amount of energy additional to this as determined by the integral of the voltage drop across the thermistor and current through it during the turn on period (both of these are changing dynamically).

Assuming those sources are correct, can anyone explain why the energy stored in the capacitor is equal to the energy dissipated through the resistor?

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The energy that the battery supplies is the integral of the \$V_{in}I(t)\$, which is just the total charge that left the battery during the charging time:

\$E_{battery} = V_{in} \cdot \int{I \cdot dt} = QV_{in}\$

All that charge got to the capacitor (since they share the same current), and the energy in the capacitor can also be written in terms of charge (using Q=CV):

\$E_{capacitor} = \frac{1}{2}CV_{in}^2\ = \frac{1}{2}QV_{in}\$

Subtracting the two will give the amount of energy lost to the resistance during charging:

\$E_{battery} - E_{capacitor} = E_{resistance} = \frac{1}{2}QV_{in}\$ which is the same as the energy in the capacitor.

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  • \$\begingroup\$ Ahhhh thanks! Pretty simple once you know whats happening, but a very interesting result! I guess from this you can infer that for any capacitor charged from a fixed DC source, you always waste just as much power charging it as you store in the capacitor! \$\endgroup\$ – gbmhunter Jan 13 '15 at 1:56
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The answer by caveman is certainly correct, but doesn't explain why it is so. Let me try to explain it with words. It's quite simple, actually.

The circuit is a series arrangement, so at any time the power partition is the same as the voltage partition.

  • The voltage across the cap is proportionnal to the charge that was already delivered (Vcap = Q/C).

  • The voltage across the resistor makes up for the rest of the battery voltage.

So, at 10% charge, the voltage split is 10%/90% for the cap/resistor. At 90% charge, the voltage split is opposite (90%/10%). Same goes for power split.

If you take both into account, the same energy has been delivered to the resistor (mostly at the beginning) and to the capacitor (mostly at the end).

Finally, please note that :

  • this even split of energy is true only if the capacitor is initially fully discharged. If it's partly charged, more energy will go to the capacitor than to the resistor.

  • if the capacitance varies with the voltage, as is usually the case, this even split is not true.

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  • \$\begingroup\$ That's also quite a nice explanation. I also like to think about it as "If you make the resistance smaller and smaller, it's still dissipates the same amount of power because the current gets larger and larger, and \$ P_{res} = I^2 R \$. \$\endgroup\$ – gbmhunter Jan 14 '15 at 6:30

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