The discussion centers on the relationship between voltage and energy storage in capacitors, specifically addressing the 1/2 factor in energy calculations. When a capacitor is charged by a battery, the voltage across the capacitor can equal the battery voltage, yet the energy stored is only half of what is supplied due to energy losses in the charging process. This phenomenon is explained through the concepts of internal resistance and the exponential nature of charging, which results in energy dissipation as heat. The key takeaway is that while voltage represents potential energy, the actual energy stored in a capacitor is calculated as CV²/2, where C is capacitance and V is voltage.
PREREQUISITESElectrical engineers, physics students, and anyone involved in circuit design or analysis will benefit from this discussion, particularly those interested in energy storage mechanisms and capacitor behavior.
In order to charge a capacitor, you need an Energy source (battery etc). All sources of Electrical Power have an internal resistance which is in series with the output. Every Coulomb of charge that goes into the Capacitor will require Energy to get through the resistor. This amount of energy is the same as the energy that's stored, however low the series resistor happens to be; it will just happen quicker for a low resistor value. Whilst the Voltage across the Capacitor is less than V, current will keep flowing from battery to Capacitor. The value of the current flowing will reduce exponentially with time until equilibrium is reached. The final value must consequently end up the same as the battery volts. It's an exponential change, so it never 'really' gets there - as with heating and cooling curves etc.. But it's 'near as dammit' within a few RC 'time constants' (look it up).JFS321 said:
You cannot always expect to get an intuitive answer to this sort of question. There are a number of 'proofs' for it and they are based on Calculus or a simpler, verbal version of calculus. This link gives a fairly rigorous argument for you.JFS321 said:
Half of the energy supplied by the battery gets "lost" in the resistance of the charging path as heat energy(i2Rt ).JFS321 said:
.Try the analogy of comparing an electric motor and a spring. The electric motor is providing its designed power all the time but the spring is delivering less and less as it runs down. In well designed systems, clockwork motors (and supercapacitors used as energy reserves) are not allowed to run down completely but operate from 'full' to, perhaps 'half full'. By operating near the 'top end' they will be delivering much more constant power than if they were allowed to run down completely. This can be compensated for with suitable technology. I have an ancient (wind-up) Wall Clock that has a Fusee movement. The spring drives a tapered drum, with a light chain that operates on a varying diameter so that the tension is the same over the range of energy that the spring is allowed (26 turns of the key, so it happens). The clock stops running long before the spring is exhausted but whilst it is going, the actual power delivered to the movement is kept nearly constant, which provides more accurate timekeeping. Damned clever, those Victorians!JFS321 said:
sophiecentaur said:
Whatever the value, half of the energy is lost in it. Draw a circuit with a voltage source, a switch, a series resistor and a series capacitor. Then calculate the energy dissipated in the resistor when the switch is connected. It will be half the total energy supplied and equal to the stored energy in the capacitor. It's only the RC time constant that will be affected by the value of R. It's all in the top half of the 'triangle' graph.David Lewis said:
There will definitely be electro magnetic radiation from the circuit while the current is changing. This can be detected as radio wavessophiecentaur said: