# how the capacitor equation can be modified?

I have a 500 mF capacitor with a charge of 1 V and connect an 1 ohm resistor in series.

What will be the maximum current flowing through the resistor? It may be 1A, right?

But the max capacity of this capacitor at 1 V is 500 mC, so how can we say that in our circuit the max current that will flow is 1 A from equation:

$$I = I_{max} \cdot e ^ {-{{t}\over{RC}}}$$

$$I_{max} = {{Q_{max}}\over{RC}}$$

• There are holes in your understanding of calculus. – Ignacio Vazquez-Abrams Oct 4 '16 at 15:07

When t=0, you get I=I(max) * e^0 e^0 == 1, because any number to the power of 0 equals 1.

So you get I_0 = Imax * 1 = Imax = Q/RC = 0.5/(1 * 0.5) = V/R = 1A, of course the current falls as the cap discharges (captured by the e^-t/rc terms as t increases), but at t=0, current is clearly 1A.

I see no problem here.

• Thank u sir, But capacitor can only hold up 500m Columbs so how we will get 1 amps from it ? – parth Oct 4 '16 at 15:27
• @parth: How can a car go 90kph when the tire is only 83cm in diameter? – Ignacio Vazquez-Abrams Oct 4 '16 at 15:45
• Okay sir i understood. – parth Oct 4 '16 at 15:54
• But if car has only the capacity to go with 80kmph so how can it will go with speed 90kmph ? – parth Oct 4 '16 at 16:00
• More like, how can a car go 90km/h if it only has fuel for 45km range in the tank.... Charge is current * time, make the time small enough and very little charge is consumed even at large currents. In this case we ask what the current (equivalent to the voltage because 1 ohm resistor) is at any given point in time, do the calculus. I=V/R = q/RC. dq/dt = I, initial condition is q = 500mC. – Dan Mills Oct 4 '16 at 16:20

Regardless of the amount of energy it's storing, if a capacitor is disconnected from its charging source and a 1 ohm resistor is suddenly connected across the capacitor when the voltage across the capacitor is 1 volt, the instantaneous current into the resistor will be 1 ampere for a nearly infinitesimal amount of time.

The reason for that is that since the voltage across the capacitor is equal to Q/C, the passage of even a single electron into and out of the resistance will result in work being done, Q being diminished and, since $V=\frac {Q}{C}$, V being diminished as well.

• Thank u for answering sir. Can u plz give me more easy explanation for it. – parth Oct 4 '16 at 16:55
• @parth: Sure. Let's say you have a vertically oriented tube which can hold 1000 marbles which weigh one gram each, that the tube contains 1000 marbles, that there's a scale at the bottom of the tube which can weigh the column of marbles, and that there's a valve, anywhere, which can open and close quickly enough to capture a single marble. If the valve removes one marble from the column, what will the scale register? – EM Fields Oct 4 '16 at 18:25