# Power calculation

Let's say I'm gonna charge a 1mF capacitor from 0 to 100V in 10 seconds. Assuming there's no limiting resistor nor any other losses, the total energy that must be tranfered from the 100VDC supply to the cap is:

$E&space;=&space;1/2*C*V^2&space;=&space;5J$

So, the power transferred in 10 seconds:

$P&space;=&space;E/t&space;=&space;0.5W$

Ok. The charge needed to charge the 1mF capacitor from 0 to 100V is:

$Q&space;=&space;C&space;*&space;V&space;=&space;0.1C$

So, the power transferred in 10 seconds is:

$P&space;=&space;V&space;*&space;I&space;=&space;V&space;*&space;Q/t&space;=&space;1W$

Which one is the correct? What did I do wrong?

• Can you take a look at this answer and let us know if it clarifies things? If not, can you indicate the specific doubts that arise/remain? Jul 25, 2020 at 23:14
• V in your final equation cannot be constant over the 10 second period. If the current is constant, for example, V will be increasing linearly, and the average voltage will be V/2, hence power is VI/2.
– Chu
Jul 25, 2020 at 23:49
• What value are you using for V in the last equation and why? (@chu: Snap!) Jul 25, 2020 at 23:50
• It's clearer now, thanks! Jul 27, 2020 at 13:39

Let's say I'm gonna charge a 1mF (1000 μF) capacitor from 0 to 100V in 10 seconds

That can be achieved with a constant current circuit in series with the capacitor fed from the 100 volt supply. The formula that defines the current is this: -

$$I = C\dfrac{ΔV}{Δt}$$

So, ΔV is 100 volts and Δt is 10 seconds. That means that I = 10 mA for a capacitance of 1000 μF.

From the perspective of the power supply, it is generating 100 volts and there is a load taking 10 mA, That's a constant power of 1 watt over the ten second period. If we converted power to energy (joules) by multiplying by time we get 10 joules.

But we also know the stored energy (W) of the capacitor when it reaches 100 volts: -

$$W = ½\cdot C\cdot V^2$$

Plugging the numbers in we get 5 joules. This means that 5 joules is converted to heat inside the constant current circuit.

Assuming there's no limiting resistor nor any other losses

That means you have made a contradiction in your question because you can't have a capacitor charging over ten seconds without some form of current limiting (be it a constant current source or a plain ordinary resistor). That current limiting consumes power. Neither can you instantly apply 100 volts to your capacitor because that implies infinite current and nobody has infinite current.