Suppose we have a capacitor of capacitance $C$ and a cell of emf $E$, why is the maximum energy stored in a capacitor equal to $\tfrac{1}{2}CE^2$? I feel confused because, the potential difference against the capacitor will be less than the emf of the cell because of the potential drop across the resistor. What is the flaw in my thinking?
[Physics] Energy stored in a capacitor in an RC circuit
capacitanceelectric-circuitselectrical-resistance
Related Solutions
Here, initially some current flows through R1 and C1 but only until C1 gets fully charged once C1 is fully charged no current flows through the resistor.
If you charge up a capacitor through a resistor current will flow until the voltage across the capacitor is the same as the source. This is an exponential process and never really gets there but it comes pretty close after some time.
When this happens current doesn't flow through the resistor That means the voltage drop across the capacitor is equal to the EMF of the cell.
Energy stored in a capacitor would be = 1/2*QV or 1/2*C*V^2. (V is now the EMF of battery, C is the capacitance of the capacitor, Q is the charge on the capacitor.)
Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current.
If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the switch is closed. If you introduce a very small resistance in the charging path the current will be large and large losses will occur despite the resistor's small resistance. As you increase the resistor the current will decrease but the increasing R causes increased power loss.
The closest real world approximation to lossless capacitor charging is to use a buck converter.
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Best Answer
Once the capacitor has fully charged the current in the circuit will be zero, so the voltage drop across the resistor is zero and hence the voltage across the capacitor is equal to the cell voltage.
Having said this, the current falls exponentially with time so in principle the current takes an infinite time to fall to zero, and the voltage across the capacitor takes an infinite time to rise to $E$. However this is a somewhat pedantic position as in most cases the voltage across the capacitor will rapidly be indistinguishable from $E$.