Explain charging and discharging of capacitors and explain time constant.?

Question

I have been given this question in a Electronics assignment which has to be attempted for three marks.Can u give me a suitable answer(I am in XI standard).Also what actually is a capacitor?

Answer 1

See ref. 1. This is an excellent explanation.
About exponential functions and time constants (from another answer I wrote):
An exponential function diverges from a base value by a factor of e (or converges to a base or asymptote value, reducing the difference by a factor of 1/e) for each time interval known as a "time constant". In a circuit in which a capacitor C is charged from a voltage source V through a resistor R, the time constant T = R*C. The voltage on the capacitor Vc approaches V as a function of time t as follows: Vc = V * (1-e^(-t/T). This equation represents an asymptotic or "decay" function (even though increasing) since the power of e is negative. What this means is that in each successive time interval of T, the difference between Vc and V decreases, being multiplied by a factor of 1/e or about 0.37.
Conversely, if C is charged to V and you discharge it through R, the voltage approaches zero asymptotically in the same manner: Vc = V * e^(-t/T).
You may already know about "e"; it's the base of the natural logarithms and is a factor in almost every kind of natural growth (if unlimited) in which the rate of growth is proportional to the current value (and the exponent of e is positive), and natural decay in which the rate of decay is proportional to the difference between the current value and the limit the value is approaching. Note that you will often find reference to a "rate constant", or growth or decay constant, which is the same as the reciprocal of T used in the equations above, so you'll see terms like e^-k*t instead of e^-t/T. Ref. 2 has more about exponential growth and decay.

Answer 2

it is quite no longer a distinct rule. purely imagine of the discharging case as having a voltage source of 0V. Charging: source + resistor + capacitor = 0 source = - (resistor + capacitor) You have a tendency to brush aside the signal because you may parent out which way modern-day flows intuitively. Discharging: 0 + resistor + capacitor = 0 resistor = - capacitor lower back you may ignore with regard to the signal in case you understand that modern-day flows the different direction of at the same time as it change into charging.