Hello,
For learning electrical engineering there is importance of analog please tell where to start analog engineering?
for example we have charge 2200uf capacitor of rating 50V
but i applies a 12v source.
and there is load R how to find time it will store charge?
You can start by learning physics and math. Learn about the passive components used in electronics: resistors, capacitors, and inductors. Learn how resistors behave in circuits by learning to use Ohm's Law and Kirchoff's Laws to analyze the currents in, and the voltage drops across, resistors in increasingly complicated
series and parallel circuits with one or more voltage and/or current sources. Learn about voltage and current sources, both DC as well as AC, and their Thevenin and Norton equivalents. Learn how capacitors store and release energy. Learn how inductors store and release energy. Learn basic calculus skills of differentiation and integration. Learn how to examine a passive circuit and set up and solve differential equations describing the currents and voltage drops in the circuit.
Once you have mastered circuit analysis using passive components you can begin a course of study in analog circuit engineering using active components. Start with the bi-polar junction transistor (BJT) and master it's biasing and use in common-emitter, common-base, and common-collector circuits. Move on to field-effect transistors (FETs) and do the same thing for common-source, common-gate, and common-drain circuits. This should keep you busy for the first year. Do a lot of lab work, measuring voltage and currents in actual (real) circuits to compare the measured results against your circuit analysis.
Avoid using SPICE simulators until you fully understand how circuits work. Simulators don't teach you anything. They can be a big help in performing "what if" analysis, such as "what happens if I change this resistor value?" But you need to understand the circuit first. Like spreadsheet programs, there is a lot of hidden processing going on in a simulation program. Unless you understand the models used in the simulation, and the limitations of the processing, simulator results can lead you down a prim-rose path to disaster.
i(t) = C dv(t)/dt
what is this dv?
It is derived from the equation that
defines capacitance:
C = Q/V, where C is the capacitance in farads, Q is the charge separation on the capacitor in coulombs, and V is the voltage across the capacitor in volts. Assuming the capacitance is constant, you can write this as
Q = CV and differentiate both sides with respect to time to get dQ/dt = C dV(t)/dt. But dQ/dt (the derivative of charge with respect to time or the rate of change of charge) is a definition of current, I(t), as a function of time. So by substitution you get I(t) = C dV(t)/dt.
Looking at the right side of this equation, if dV(t)/dt is a constantly increasing voltage (a ramp) a constant current will be charging the capacitor. Or, dividing both sides by C to get I(t)/C = dV(t)/dt you will find that if I(t) is a constant current, the voltage across the capacitor will be a linearly changing ramp. The capacitor acts as an integrator of the current input to the capacitor.
In your first post, you asked how long a capacitor will store charge if a load, R, is attached to the capacitor. The load will continuously discharge the capacitor until (eventually) no charge remains. Theoretically, this takes forever, but the discharge current decreases exponentially as the capacitor voltage decreases. Using calculus and circuit analysis you can derive the equation for the voltage, Vcap, remaining across a capacitor, C, with parallel-connected resistor, R, initially charged to voltage Vinit. It is Vcap = Vinit [e^(-t/RC)], where "e" is the base of natural logarithms (about 2.71828) raised to the negative power of the quotient t/RC, where t is the elapsed time of the discharge. If you plug in values for t, R, and C and perform the calculation, you will see that Vcap approaches zero, and is nearly zero, when t = 5RC. This is the "practical definition" of "fully discharged" but in reality the discharge continues as long as the resistor is connected, until eventually the voltage across the capacitor is just noise... the noise voltage being generated thermally by the discharge resistor.
See Johnson Noise for more information on that.
Good luck with your studies!