well guys i am quite nooby at electronics but very enthusiastic to learn, i have got couple of questions about capacitor would be grateful to share the answers if u know them.

for the questions below just assume a very simple circuit DC power supply V, connected to a capacitor C and then goes to the ground.


1- how long does it take for a capacitor of C, connected to a DC power supply of V to be charged ?

2-i know the current cant pass through the capacitor with DC power supply, but lets say the capacitor has been fully charged after the time t, then we wanna discharge the capacitor, so how we can do it by discconecting the main DC source right ?

3- so can we say after time t the charged capacitor has obtained the same voltage as power supply V ?

4- how is it possible to calculate the current provided by the charged capacitor obviously we know the charge Q, and the time t is keep changing so it means the current also keep changing i mean going down cuz as the time increases the currect should decrease I=q/t right ?

thanks guys for ur time
cheers

 

Your questions have to take into account a resistance somewhere. Otherwise (on paper), the current flow from the power supply would be infinite for zero time and the capacitor would charge instantaneously. Obviously, this can't happen in the real world. The supply will have some resistance, the capacitor will have some resistance, and the wires connecting them will have some resistance. Even so, it is physically possible for them to be so low that extremely high, even dangerously or destructively high, currents can result.

But let's assume that the total resistance in your circuit, from all sources, is R and that it is high enough to keep bad things from happening.

When the capacitor is connected to the power supply through a resistor, the voltage across the resistor will be (Vs - Vc) where Vs is the supply voltage and Vc is the capacitor voltage. So the current will be (Vs-Vc)/R and this current will either be charging or discharging the capacitor, depending on which direction it is flowing.

When you first connect a discharged capacitor to the supply, Vc will be zero and the current will be a maximum. But as charge accumulated Vc will start to grow and the current will decrease and the rate at which the capacitor voltage changes will decrease.

The result is that the capacitor will charge according to an exponentially decaying curve. The rate is contolled by the "time constant", which is the product of R and C, so

tau = RC

At the end of one time constant, the capacitor will be charged to ~63% of its final value. After a second time constant, the capacitor will charge ~63% of the remaining way to the final value. Each additional time constant will see the voltage move another ~63% of the way to the final value. In theory, it will never actually stop charging, but after 5RC it will be more than 99% of the way and this is when we typically call it "charged".

Once it is charged, if you disconnect the supply then the capacitor will NOT discharge. It will sit and hold the charge for quite some time. Physical capacitors have a leakage resistance that will, over time, result in the charge finding its way from one plate to the other and, thus, discharge the capacitor. But this can happen over the course of minutes to months (or longer). To discharge the capacitor, you want to connect an external resistor between the terminals. Note that this is identical to the prior case, except that Vs now happens to be 0V. So the same process and equations apply.

To go much further, I need to know what your math background is like and, in particular, do you have any calculus yet?

 

Click on my name and look at my blogs. One of them has the equations for capacitance charging time.

 

Your questions have to take into account a resistance somewhere. Otherwise (on paper), the current flow from the power supply would be infinite for zero time and the capacitor would charge instantaneously. Obviously, this can't happen in the real world. The supply will have some resistance, the capacitor will have some resistance, and the wires connecting them will have some resistance. Even so, it is physically possible for them to be so low that extremely high, even dangerously or destructively high, currents can result.

But let's assume that the total resistance in your circuit, from all sources, is R and that it is high enough to keep bad things from happening.

When the capacitor is connected to the power supply through a resistor, the voltage across the resistor will be (Vs - Vc) where Vs is the supply voltage and Vc is the capacitor voltage. So the current will be (Vs-Vc)/R and this current will either be charging or discharging the capacitor, depending on which direction it is flowing.

When you first connect a discharged capacitor to the supply, Vc will be zero and the current will be a maximum. But as charge accumulated Vc will start to grow and the current will decrease and the rate at which the capacitor voltage changes will decrease.

The result is that the capacitor will charge according to an exponentially decaying curve. The rate is contolled by the "time constant", which is the product of R and C, so

tau = RC

At the end of one time constant, the capacitor will be charged to ~63% of its final value. After a second time constant, the capacitor will charge ~63% of the remaining way to the final value. Each additional time constant will see the voltage move another ~63% of the way to the final value. In theory, it will never actually stop charging, but after 5RC it will be more than 99% of the way and this is when we typically call it "charged".

Once it is charged, if you disconnect the supply then the capacitor will NOT discharge. It will sit and hold the charge for quite some time. Physical capacitors have a leakage resistance that will, over time, result in the charge finding its way from one plate to the other and, thus, discharge the capacitor. But this can happen over the course of minutes to months (or longer). To discharge the capacitor, you want to connect an external resistor between the terminals. Note that this is identical to the prior case, except that Vs now happens to be 0V. So the same process and equations apply.

To go much further, I need to know what your math background is like and, in particular, do you have any calculus yet?

Thank u so much for ur answer i do appreciate it i have quite strong math background i took calculus, know all about differentiation,intigeration,transforms [la place, z series, fourier],...
the only question left for me is, Vc and I,,, V and and R given and i know Vc is a function of I, but is there anyway to calculate Vc or I, Vc could be determined using C=Q/V but what about Q, can we say I=Q/t.
Thank u so much for ur help i do appreciate,
regards,
shayan

 

Thank u so much for ur answer i do appreciate it i have quite strong math background i took calculus, know all about differentiation,intigeration,transforms [la place, z series, fourier],...
the only question left for me is, Vc and I,,, V and and R given and i know Vc is a function of I, but is there anyway to calculate Vc or I,

Vc could be determined using C=Q/V but what about Q, can we say I=Q/t.

Saying I=Q/T gives you the average current over a period of time T. But to find v(t) you need the instantaneous current, i(t), which dQ/dt.

So the constitutive equation for a capacitor is:

Q = CV,

hence

i(t) = dQ/dt = C dv(t)/dt

 

Cool thank u so much now everything makes more sense
have a great day,
once again thanks
cheers