We know investigation how to discharge a capacitor while exerting some control on how the discharge time will be. You can of course place a lead directly across a capacitor to discharge it very quickly and possibly cause a visible spark.
For large capacitors such those in Tv sets this procedure should not be attempted because of the high voltages involved unless of course you are trained in the maneuver.
In a second contact for the switch was added to the circuit in permit a controlled discharge of the capacitor. With the switch in position 1, we have the charging network described in the last section. Following the full charging phase if we move the switch to position 1, the capacitor can be discharged through the resulting circuit in the voltage across the capacitor appears directly across the resistor to establish a discharge current. Initially the current jumps to a relatively high value then it begins to drop. It drops with time because charge is leaving the plates of the capacitors which in turn reduces the voltage across the capacitor and thereby the voltage across the resistor and the resulting current.
Before looking at the wave shapes for each quantity of interest note that current ic has now reversed direction as shown in parts the voltage across the capacitor does not reverse polarity but the current reverse direction . We will show the revers on the on the graph. In all the waveforms in the mathematical expression use the same e-x factor appearing during the charging phase .
For the voltage across the capacitor that is deceasing with time the mathematical expression is:
Vc=Ee-l/t
For the circuit the time constant pri is defined by the same equation as used for the charging phase.
That is ,
pri=RC
Since the current decreases with time , it will have a similar format,
Ic= (E/R)e-l/t
For the configuration since Vr=Vc the equation for the voltage Vr has the same format,
Vr=Ee-l/t
important tropics............
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