If a charged capacitor is connected to a current source and resistor as shown above, the resistor is constantly consuming 0.001J/s. If you were to make an energy-time graph of the capacitor, you would see a straight linear decrease.
These equations hold for a capacitor:
\$ I = C \frac{dV}{dt} \$ (1)
\$ V = I\frac{t}{C} + D \$ (2) <= this is a linear decrease of voltage
Thus, because I is also constant, you would see a straight linear decrease in a voltage-time graph as the capacitor is discharging.
However, the energy of a capacitor is
\$E = \frac{1}{2}CV^2\$ (3)
and the voltage is
\$ V = \sqrt{2 \frac{E}{C}} \$ (4)
If the energy inside the cap is decreasing linearly, how can the voltage be also decreasing linearly according to that equation?
Best Answer
In your circuit, the current source is absorbing or supplying energy, so the resistor is not the only place the capacitor energy can be transferred to. When the capacitor is charged above 0.1 V, the current source will be absorbing energy, and when the capacitor charge is below 0.1 V, the current source will be supplying energy.
If you consider the energy absorbed by both the current source and the resistor, they will add up to the energy being discharged from the capacitor.
This is not correct.