Capacitor related query

Question

Why does a capacitor charge only upto the voltage of the source? I mean in an r.c. circuit if we have a resistor that is in series with the capacitor then because of the potential drop there will be less amount of current flowing to the capacitor. So shouldn't its final steady state voltage value be equal to the potential drop across the resistor? And on the capacitors its written 63 volts. So I guess it can withstand a voltage of that value. So if its steady state voltage is not equal to the potential drop across the resistor then why isnt it equal to the max value that is 63 volts. I know its a stupid doubt, but any help will be appreciated. The voltage of the source is 12 volts.

Answer 1

Why does a capacitor charge only upto the voltage of the source?

Step by step:

(1) When the capacitor voltage equals the source voltage, the voltage across the resistor in the series RC circuit is zero

(2) By Ohm's Law, the current through the resistor must be zero too.

(3) Because it is a series circuit, if there is zero current through the resistor, there is zero current through the capacitor.

(4) Since there is zero current through the capacitor, the rate of change of the capacitor voltage is zero.

Thus, if the capacitor voltage equals the source voltage, the capacitor voltage cannot be changing.

If the capacitor voltage cannot be changing when the voltage equals the source, how could the capacitor voltage change to be greater than the source?

Answer 2

A capacitor is storage. it has a capacity for what current goes in and stores it. I need the terminology to put it into words but it would be like saying you cant get out what you didn't put in the amount depends on the size of the capacitor.

EG if you have a bucket that can only accept or give 1 nickel at a time and you put in 50 nickels then it will only give you a nickel at a time 50 times.

I know it is more complex than that but that is the simplest way I can explain that anyone can understand.

With Capacitors we are talking volts and amps volts being the strength and amps being the capacity.

Sometimes looking at things from another angel broadens perspective, correction, always.

Answer 3

Here's how I would convince myself of the correct answer.

1. Draw a circuit diagram showing the voltage source, resistor, and capacitor. (I assume it's in a simple series circuit?)

2. Next, write out Kirchhoff's loop rule. You should find something like $V_\text{source}-V_\text{resistor}-V_\text{cap}=0$. Note that this equation is true at any time, not only for the steady state solution.

3. Finally, in steady state we have $I=0$. Knowing that the current is zero will help you figure out that one of the voltage terms in your equation takes on a particularly nice form. You can now solve for $V_\text{cap}$

My guess is that $63\ \text{Volts}$ listed on the capacitor itself is indeed the maximum potential difference that one should apply between the leads of the capacitor. It does not mean that the capacitor will automatically take on that value once "fully charged."