How current relates to capacitance

[JFS321]

Hi all,

I have a simple circuit with a battery, resistor, and finally, the capacitor.

Am I correct in saying that adding additional resistors will increase the charging time, because adding resistors will decrease the overall current?

Thanks--

 

[technician]

Basically...yes...with the information you have provided!

PS...unless you connect the resistors in parallel

 

[sophiecentaur]

Hi all,

I have a simple circuit with a battery, resistor, and finally, the capacitor.

Am I correct in saying that adding additional resistors will increase the charging time, because adding resistors will decrease the overall current?

Thanks--

It is really far more fruitful to describe what happens using Maths.
We assume the three are in series. The time constant (time for the voltage across the Resistor to go to 1/e of any initial value) is given by RC. It's an Exponential variation. As the Voltage decreases exponentially, so does the current flowing because I = V/R.

 

[technician]

A more direct way for you to consider the time constant is that RC is the time taken for the CURRENT in the series circuit to decrease to 1/e of its initial value.
(your question related to the current)
T = RC so larger R means longer time for the current to decrease.
Increasing R decreases the current which increases the time to charge the capacitor.

 

[sardar]

Hello,

Yes, you are correct in saying that adding additional resistors will increase the charging time of a capacitor. This is because resistance and capacitance are inversely proportional to each other. As resistance increases, the current decreases, and therefore, the rate at which the capacitor can charge also decreases. This relationship can be mathematically represented by the equation Q=CV, where Q is the charge on the capacitor, C is the capacitance, and V is the voltage. As you can see, if resistance increases, the voltage stays the same, but the capacitance decreases, resulting in a longer charging time.