Capacitor
RC CIRCUIT - BASICS
What happens???
What happens when we connect a resistor in series with a capacitor???
Let us assume an $1A$ rating wire is connected to a capacitor of $1F$ and a battery of $1V$. The above capacitor can charge upto $1$ coulomb. $$C=\frac{Q}{V}$$$$Q=CV=1$$
Since we are using $1A$ rating wire, $$I=\frac{Q}{t}$$$$t=I×Q=1\;s$$We can say, it will take $1sec$ for the total charge to flow through this wire.
Now assume a resistor of $1k\Omega$ is connected in between the battery and the capacitor.$$I=\frac{V}{R}$$$$I=\frac{1}{1×10^{3}}$$$$I=1×10^{-3}$$$$I=1\;mA$$So, this resistor will limit the flow of current through the capacitor. Hence$$t=\frac{Q}{I}$$$$t=\frac{1}{1\;mA}$$$$t=1000\;secs$$it will take approximately $1000\;secs$ for the total charge to flow through the resistor.
From the above discussion, we can say connecting a resistor in series with the capacitor will slow down the charging-discharging process. This slow down is technically termed as RC delay.
When we connect a resistor in series with a capacitor, it will slow down the charging-discharging process.
Water-tank Analogy:
Assume a larger width pipe is connected to a water tank to drain the water.
If a smaller width pipe is connected in between the large width pipe, the draining process will slow down.
Now assume,
- Larger width pipe = conductor (wire)
- Smaller width pipe = resistor
- free electrons = water
- Tank = capacitor
We can now understand how a resistor affects the charging-discharging process.