Why is current drawn in an RC circuit (in a circuit powered by DC voltage supply) independent of the capacitor used?
While the capacitor is charging current drawn from the battery only depends on resistor and on the value of capacitor.
[Physics] Current in RC circuit
capacitanceelectric-circuitselectric-currentelectrical-resistancevoltage
Related Solutions
The circuit is drawing infinite amount of current from the battery. Have I correctly understood this ?
Yes.
There will be a potential difference across the resistor in parallel to capacitor and that potential difference will be resposnsible for charging it
The potential across the capacitor can't change instantaneously.
Therefore in the time immediately after the switch closes, the voltage across the resistor (the one in parallel with the capacitor) is zero. From Ohm's law, then, there is no current through this resistor in that instant.
To find the current that is charging the capacitor (in the instant immediately after closing the switch), you can use KCL at the node where the capacitor and the two resistors are all connected.
Alternately, you can replace the voltage source and the two resistors with a Thevenin equivalent circuit, and again find the charging current as time progresses after the switch is closed.
Since this is a physics q and a, a physics explanation is in order.
There are two kinds of current.
Conduction current is a net flow of charges. It is was people usually think of when the word "current" is used
Displacement current is another form of current, first recognized by Maxwell. Displacement current plays an essential role in Maxwell's equations. Displacement current density is proportional to the time derivative of the change of electric flux density.
When electron current flows into one side of a capacitor, the electrons accumulate, as there is no place for them to go. As the electrons accumulate, the electric flux density changes. This causes, or perhaps "is" a displacement current.
On the opposite plate of the capacitor, a similar process occurs, but with opposite electrical polarity.
The displacement current flows from one plate to the other, through the dielectric whenever current flows into or out of the capacitor plates and has the exact same magnitude as the current flowing through the capacitor's terminals.
One might guess that this displacement current has no real effects other than to "conserve" current. However, displacement current creates magnetic fields just as conduction current does.
This answer is perhaps more than one might want to know, but it is part of the story of electricity that is worth telling.
Best Answer
In a charging RC circuit where the capacitor is initially uncharged, the charges will move as if the capacitor is essentially absent. Therefore, the initial value of the current is just equal to $V/R$.
If the RC circuit starts with a fully charged capacitor and is discharging, then once the current starts the capacitor acts like a battery. The circuit is then essentially a resistor in series with a battery, and the initial current is once again $V/R$ (since the capacitor's initial potential difference has a magnitude of $V$).
At times when the current is changing, the value of the capacitance does effect how quickly the current changes with time constant $RC$. This is because the amount of charge stored on the capacitor changes the potential drop across the capacitor due to $V=CQ$. The more charge that is present, the more it “fights the battery” in the charging case, and the more it pushes charge of itself in the discharging case.
In either case, the magnitude of the current is given by $$I(t)=\frac VRe^{-t/RC}$$ so the current does depend on the capacitance, but the initial current does not.