# Does the emf source live forever with a purely capacitive load?

Pardon me, but this question has been troubling me for several days. I've searched a lot but couldn't find any convincing explanation.

In a purely resistive circuit, instantaneous power is given by $P = V_m\sin(\omega t) \times I_m\sin(\omega t)$. This power heats up the resistor and is always positive, the energy is flowing from emf source to the resistor; so the emf source will die after transferring all its energy to the resistor.

In a purely capacitive circuit, $P = V_m\sin(\omega t) \times I_m\sin(\omega t+90) = \frac12 V_mI_m\sin(2\omega t)$. Average value of this is $0$. This means energy goes back and forth between capacitor and the emf source.

Does the emf source live forever? If the emf source were ac mains, would I get any bill? Since I didn't use any energy, what was working on charging and discharging the capacitor? Charging and discharging requires work and needs energy right?

Charging and discharging requires work and needs energy right ?

It depends. In the real world, losses always occur. So, yes.

In an ideal world, with a lossless capacitor, the energy you get back from discharging exactly balances what you put in to charge it. So you could run charging and discharging for ever with no loss of energy.

If the emf source were ac mains, would I get any bill ?

It depends on what sort of customer you are. In the UK, domestic customers are only charged for real energy used, so if the meter was working properly, no, you would not get a bill. Industrial customers are charged for reactive VA as well, as an incentive for them to get their power factor reasonable.

• Gotcha! Ty :) so if a domestic customer had hooked a capacitor to the mains directly and let it charge and discharge like crazy, assuming losses were negligible, he wouldn't have to pay anything. When you refer to losses, are they due to the reactance $1/(j\omega C)$ ? This can be huge as the numerical value of $C$ is usually small ? – Hiiii Jun 8 '17 at 10:17
• No, when I, and most electronic engineers, refer to losses, we are talking about real power, heat dissipation, resistance of conductors generating I2R as heat. Pure (ideal) reactance, 1/wC and wL, generate no heat, though they can cause large currents to flow which do cause heat losses in resistances elsewhere. This is why power utilities charge factories for drawing large reactive currents, because they cause large power losses in the supply lines and transformers. – Neil_UK Jun 8 '17 at 11:00
• Ohk.. If I understand correctly, losses in electrical engg almost always refer to i2r losses dissipated in resistors / parasitic resistances as heat. We use reactances $1/\omega C$ and $\omega L$ for computing voltage drop, currents etc. Ofcourse $P = v(t)\times i(t)$ gives the total power which includes both resistive and reactive components. I wonder how the meter separates these components(cont). – Hiiii Jun 8 '17 at 13:02
• I think I can see how the math works, thanks to my complex analysis class, but I haven't any clue how the circuit inside the meter is designed. I think taking the average value of the total power gives the real power... A simple integrator circuit might do. But reactive power looks tricky... I'll google and get back. You're great at teaching difficult things in a simple manner! Thank you so much(: – Hiiii Jun 8 '17 at 13:04
• The circuit inside a 'proper' meter digitises both voltage and current, computes instantaneous power sample by sample, and integrates to get real power. The circuit in an approximate meter computes the rms of each of the voltage and current waveforms, and the phase angle between them, and then uses cos(angle) to estimate the real power. The mechanism in the old-style mechanical meter uses voltage and current coils to drive the disk, the effect is to form the instantaneous product which is the power to drive the disk. – Neil_UK Jun 8 '17 at 14:09