Lets try to under stand how capacitor work s. Lets say we have following circuit

So lets take a capacitor and try to charge it. For that we will be needing , a voltage source , a resistor which will help us to determine how quickly or slowly our capacitor will get charged.

We need to notice that as soon capacitor start charging , the potential difference b/w the battery and capacitors start decreasing , due to which flow of current decrease,, we can say that a capacitor is fully charged when current in that particular become zero

Lets say we have a thirty five volt battery which is connected to a capacitor through a resistor . lets say that capacitor has a value of 30 F .

A switch is introduced so that , when it si turned on , the capacitor will be connected to battery , and will be charged . how ever if we don’t want capacitor to get charged , we can simply turn it off.

 

Now lets connect a DMM to capacitor, to see that what is happening to capacitor, when it is connected to voltage source .

Now lets start examining that what is happening in following figure. As soon as the switch is turned on , charges start gathering on capacitor plates, due to which potential difference is created between terminals. With the passage of time more and more charges get accumulated on capacitor, hence the voltage stored by the capacitor also increases .

It is noteworthy that when in initial position , capacitor is not charged and has zero volts across it terminal.

Figure A

As soon as the switch is turned on the capacitor starts getting charged. . note the value of current and voltage in figure A

Voltage Figure A= 8 Volts

Current in Figure A = 0.003 M A

 

Figure B

With the passage of time the voltage start building up on capacitor terminal. As the potential difference between battery and voltage source reduces , the current value start decreasing

Voltage Figure B= 19.01 Volts

Current in Figure B = 0.001 M A

 

Figure C

Voltage Figure C= 34.01 Volts

Current in Figure C= 77 Amp

 

Figure D

Voltage Figure D= 34.97 Volts

Current in Figure D= 2.63mAmp

 

Figure E

Voltage Figure E= 34.99 Volts

Current in Figure E= 5.19 mA

 

Now lets draw a graph of charging of a capacitor in terms of Current and volts

Time	Voltage (V)	Current
T1	8	0.003 M A
T2	19.01	0.001 M A
T3	34.01	77 Amp
T4	34.97	2.63mAmp
T5	34.99	5.19 m Amp

Note while charging a capacitor Voltage is increasing and Current is decreasing

 

Discharging a Capacitor

Now lets connect a fan to a capacitor , through a switch., The switch will remain off in initial conditions .

Now we have two switches . the first Switch connects the voltage source to capacitor and the other connects capacitor to load.

Also notice that fan is also not working, as it is unable to get any power , due to switch in off position

 

Now the capacitor is fully charged. We will disconnect the capacitor from battery (turning off switch one)

 

Now both switch are off, mean that there is no battery connected to the circuit . However we recently charged our capacitor, meaning that it has stored energy , which can be supplied to the circuit. So if the switch between capacitor and fan is turned on , the stored energy in fan , should turn on the fan

 

To see if than fan is working or not ,Although there is battery connected to the circuit, yet still the bulb glows, as it used the enrgy from capacitor. Once the capacitor discharge , the bulb will stop glowing