Hey guys I am trying to model the performance of a hypothetical capacitor.
Capacitance depends on its geometry, but the charging and discharging of the capacitor depend on a resistive load within the circuit.
I am trying to see the performance of this capacitor when used to satisfy the energy demand of a household or power grid. However I am not sure as to how I can provide an approximate resistance value for a household or city load.
Thank you guys for your help.
Capacitance only depends on its geometry, ...
False. Capacitance is also determined by the dielectric (the insulator) between the plates and this can have a huge effect on the value obtained.
... but the charging and discharging of the capacitor depend on a resistive load within the circuit.
The load doesn't have to be purely resistive but we'll go with "resistive" for now.
I am trying to see the performance of this capacitor when used to satisfy the energy demand of a household or power grid. However I am not sure as to how I can provide an approximate resistance value for a household or city load.
Use $ P = \frac {V^2} R $ which you can rewrite as $ R = \frac {V^2} P $ where $V$ is the city supply voltage and $P$ is the demand of the city in watts (W) (and you'll have to apply the correct conversion if going from kW, MW or GW).
Energy is measured in joules (J) which are equal to watt-seconds (Ws) and 3600 Ws = 1 Wh (watt-hour). 3,600,000 Ws = 1 kWh.
The energy stored in a capacitor is given by $ E = \frac 1 2 C V^2 $. Note that the capacitor voltage will start to fall immediately you start drawing charge from it and from the formula above you should be able to see that 75% of the energy will be gone when it falls to half voltage.
You now have enough information to start doing some calculations. Try this: What value of capacitance do you require to run a 1 kW load at, say, 200 V (chosen to make the maths easy) for 10 hours?
Just checking, but you are aware that capacitor storage is DC and that city supply is AC, aren't you?
