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How does the current through capacitor and resistor network be decreasing from maximum, when the voltage across them is increasing from minimum?

And also to my knowledge the charge that's get reside the right plate of capacitor repels the electron on left part,for the flow of current. If the resided electrons on the right plate determines increasing voltage from minimum, how can the current be in decreasing from maximum?

Correct me please, if I am wrong

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    \$\begingroup\$ This question is pretty confusing. It would help greatly if you could use the built-in schematic editor and add a schematic of the "network" you are talking about, and be very clear about which voltages in the circuit you are asking about. \$\endgroup\$ – Elliot Alderson Jun 13 '19 at 20:07
  • \$\begingroup\$ Thinking about charges on plates and repulsion forces will rarely help people understand capacitors. If you have a good feel for calculus, SunnySkyGuy's answer should help you. The math works really well in electronics, trust it. \$\endgroup\$ – Mattman944 Jun 13 '19 at 20:55
  • \$\begingroup\$ Did you have any questions on my answer? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 14 '19 at 0:03
  • \$\begingroup\$ @SunnyskyguyEE75 yeah that's absolutely fine, but I want to know physics inside the concept, thank u \$\endgroup\$ – sai sri datta Jun 14 '19 at 6:08
  • \$\begingroup\$ THen understand all these concepts 1st hyperphysics.phy-astr.gsu.edu/hbase/electric/capcon.html#c1 Click on each bubble \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 14 '19 at 6:20
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I'm guessing you are describing a circuit that looks something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Assume the voltage across capacitor C1 is zero volts when switch SW1 is open (the circuit's initial condition).

$$ V_{C1}(t\le0s) = 0\,V $$

At time t = 0 seconds switch SW1 closes and an electric field passes through the circuit. When the electric field reaches C1 the potential difference between V1's '+' terminal (10V) and C1's top plate (0V) is 10 Volts.

$$ V1-V_{C1(t=0s)} = +10\,V-0\,V = 10\,V $$

Using Ohm's law we can calculate the current at t = 0 seconds as

$$ I(t=0s) = \frac{V_{R1}(t=0s)}{R1} = \frac{V1-V_{C1}(t=0s)}{R1} = \frac{10V-0V}{1\Omega} = 10\,A $$

When electrons are removed from C1's top plate, the voltage potential on C1's top plate increases (becomes more positive) relative to the ground potential. (Remember: when an electron is removed from an atom, the atom becomes a positively charged ion. So as we remove electrons from C1's top plate, the atoms in the top plate become positively-charged ions.)

At time t = 1 second the voltage on C1 has increased to approximately 6.32 Volts. The current at time t = 1 second is therefore

$$ I(t=1s) = \frac{V_{R1}(t=1s)}{R1} = \frac {V1-V_{C1}(t=1s)}{R1} \approx \frac{10V-6.32V}{1\Omega} \approx 3.68\,A $$

Note that the voltage on C1 has increased, but the current in the circuit has decreased. This is because the voltage on C1 is now "pushing back" against the voltage on V1, which slows down the flow of electrons from C1's top plate to V1's '+' terminal.

At time t = 2 seconds, the voltage on C1 has increased to approximately 8.65 volts, and the current at time t = 2 seconds is

$$ I(t=2s) = \frac{V_{R1}(t=2s)}{R1} = \frac {V1-V_{C1}(t=2s)}{R1} \approx \frac{10V-8.65V}{1\Omega} \approx 1.35\,A $$

This process continues until the electric field moves enough electrons from C1's top plate into V1's '+' terminal to cause C1's top plate potential to become 10 Volts, at which point the current stops flowing:

$$ I(3) = \frac{V_{R1(3)}}{R1} = \frac {V1-V_{C1(3)}}{R1} \approx \frac{10V-9.50V}{1\Omega} \approx 0.50\,A \\ I(4) = \frac{V_{R1(4)}}{R1} = \frac {V1-V_{C1(4)}}{R1} \approx \frac{10V-9.82V}{1\Omega} \approx 0.18\,A \\ I(5) = \frac{V_{R1(5)}}{R1} = \frac {V1-V_{C1(5)}}{R1} \approx \frac{10V-9.95V}{1\Omega} \approx 0.07\,A \\ I(\infty) = \frac{V_{R1(\infty)}}{R1} = \frac {V1-V_{C1(\infty)}}{R1} = \frac{10V-10V}{1\Omega} = 0\,A $$

So the voltage across the capacitor started at 0 Volts and increased to +10 Volts, and at the same time the current started at 10 Amps and decreased to 0 Amps.

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  • \$\begingroup\$ Thank u sir, I am able to understand by ur approach, and I have got the concept in reality \$\endgroup\$ – sai sri datta Jun 16 '19 at 9:45
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Current polarity is positive when flowing from V+ to V-.
While -ve electrons flow in the opposite direction of positive current.
This is the accepted IEE convention.

The peak current is defined by the rate of change of voltage applied limited by some series resistance including that of the capacitor, defined as ESR which is normally neglected except in high pulsed current applications.

\$I_C = C dV/dt\$

Thus current is maximum with the step voltage applied and declines with dV/dt as the capacitor voltage reaches the applied voltage.

Here shows the capacitor voltage and current. dV is 10 V but Ic is limited by 10V/1k= 10mA. When the digital sampling time is not small enough, there will be some small error in the peak transient measurement. So it reads only 9.975mA.

enter image description here

Since the voltage source is a 0 Ω "forcing function", the capacitor responds to accept the flow of current through the resistor and charge up towards the applied voltage.

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