Capacitor Discharging Phase

The discharging of a capacitor occurs when the power supply connected to the capacitive circuit is removed.
The discharging phase begins when a charged capacitor is disconnected from a power supply.
The discharging of capacitor can be controlled through resistor connected across capacitor.
We have discussed charging a capacitor. The mathematical derivations of discharging a capacitor are similar to that of charging it. The network of [Fig. 1] is designed to both charge and discharge the capacitor. When the switch is placed in position 1, the capacitor will charge toward the supply voltage. At any point in the charging process, if the switch is moved to position 2, the capacitor will begin to discharge at a rate sensitive to the same time constant $\tau = RC$.
Basic charging and discharging network.
Fig. 1: Basic charging and discharging network.
Demonstrating the discharge behavior of a
capacitive network.
Fig. 2: Demonstrating the discharge behavior of a capacitive network.
The established voltage across the capacitor will create a flow of charge in the closed path that will eventually discharge the capacitor completely. In essence, the capacitor functions like a battery with a decreasing terminal voltage. Note in particular that the current $i_C$ has reversed direction, changing the polarity of the voltage across $R$. If the capacitor had charged to the full battery voltage as indicated in [Fig. 2], the equation for the decaying voltage across the capacitor would be the following:
$$\bbox[10px,border:1px solid grey]{ v_C = E e^{-t/RC}} \quad \text{discharging} \tag{1}$$
which employs the function $e^{-x}$ and the same time constant used above. The resulting curve will have the same shape as the curve for $i_C$ and $v_R$ in the charging of capacitor. During the discharge phase, the current $i_C$ will also decrease with time, as defined by the following equation: $$\bbox[10px,border:1px solid grey]{ i_C = {E \over R} e^{-t/RC}} \quad \text{discharging} \tag{2}$$ The voltage $v_R = v_C$, and $$\bbox[10px,border:1px solid grey]{ v_R = E e^{-t/RC}} \quad \text{discharging} \tag{3}$$ The complete discharge will occur, for all practical purposes, in five time constants. If the switch is moved between terminals 1 and 2 every five time constants, the wave shapes of Fig. 3 will result for $v_C$, $i_C$, and $v_R$. For each curve, the current direction and voltage polarities were defined by Fig. 1.
Fig. 3: The charging and discharging cycles for the network of Fig. 1.
Since the polarity of $v_C$ is the same for both the charging and the discharging phases, the entire curve lies above the axis. The current $i_C$ reverses direction during the charging and discharging phases, producing a negative pulse for both the current and the voltage $v_R$. Note that the voltage $v_C$ never changes magnitude instantaneously but that the current $i_C$ has the ability to change instantaneously, as demonstrated by its vertical rises and drops to maximum values.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024
Privacy Policy