# Charging a capacitor through a high resistance

## Main Question or Discussion Point

i read from book that in order to charge an uncharged capacitor, we have to connect the capacitor and a resistor with high resistance in series to a battery with emf E. I don't understand what is the use and the effect of the resistor in this charging process? I know that the time constant is given by the product of capacitance and resistance. But is it possible to charge the capacitor without the resistor?

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Well.. charging and discharging both require a current to flow. And current will flow between two points only and only if there is a potential difference between the two points. The lead wires that we use for connections have a very low resistance.. further, these wires are assumed to have zero resistance by most of the textbooks. As such, current will not flow since there is no or a very small potential difference between the two points. To increase this difference, a resistor is added. The resistor causes a huge potential gradient in the circuit as compared to the lead wires and causes current to flow.

Another way to see this is that the amount of Energy dissipated [by Joules heating] is proportional to the square of the resistance. A charged capacitor holds some energy with it. In order to discharge, the energy must somehow be lost to the surroundings. To achieve this, we need some resistance in the circuit which will dissipate this energy in the form of heat. Higher the resistance, more is the power i.e. more energy is dissipated per unit time and hence a Capacitor will be discharged quickly.

Yes.. it is possible to charge the capacitor without a resistor. This is achieved by using a battery to create a potential difference.

Yes.. it is also possible to discharge the capacitor without a resistor by using a battery.. but it would cause the capacitor to eventually charge again.. but this time with a different polarity. For ex., if a capacitor has +q on it's right plate and -q on it's left plate, and we connect a battery [positive to left plate], then the charge of the capacitor will first go down to zero and then will again go back to q, but this time we will have -q on it's right plate and +q on it's left plate. However, it is very difficult to determine the exact moment when this charge becomes zero and disconnect it and hence we use a resistor for that purpose. This behavior may differ with rechargeable batteries [i'm not sure about that though].

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The resistor will limit the maximum current drawn from the battery.

A capacitor connected across the terminals of a battery "looks" like a dead short until the capacitor's voltage approaches the supply voltage. Using a current-limiting resistor is good practice.

Now, "can" a capacitor be charged without an extra resistor, yes, but SOME resistive element in the circuit is going to provide the current limit.

(This makes me think, what if the circuit were superconducting, what would the charging profile look like? Hmm...)

vanesch
Staff Emeritus
Gold Member
(This makes me think, what if the circuit were superconducting, what would the charging profile look like? Hmm...)
The self-inductance of the circuit will now limit the current...

Thank you very much. But i am confused that a charged capacitor will take a longer time or shorter time to discharge if the resistance in the circuit is bigger? "A larger resistance results in a smaller discharge current", is this correct?

"A larger resistance results in a smaller discharge current", is this correct?
Yes it is.

Higher the resistance, more is the power i.e. more energy is dissipated per unit time and hence a Capacitor will be discharged quickly.
You mean lower don't you?

vanesch
Staff Emeritus