1
\$\begingroup\$

A capacitor charges to 63% of the supply voltage that is charging it after one time period. After 5 time periods, a capacitor charges up to over 99% of its supply voltage. Therefore, it is safe to say that the time it takes for a capacitor to charge up to the supply voltage is 5 time constants.

Time for a Capacitor to Charge = 5RC

schematic

simulate this circuit – Schematic created using CircuitLab

Charging a Capacitor One time constant,

$$\tau=RC=(3\text{k}\Omega)(1000\mu\text{F})=3\text{ seconds, }5 \times 3=15\text{ seconds}$$

So it takes the capacitor 15 seconds to charge up to near 9 volts.

I don't understand: what if I don't attach a resistance in between? What will be the time to charge the capacitor?

\$\endgroup\$
5
\$\begingroup\$

In a perfect world, the capacitor would charge instantly. This is clear from your equation: the charge time is $$ t \approx 5RC $$ so if \$R = 0\$, then \$t = 0\$.

However, batteries are not perfect voltage sources. They have an effective resistance, which is on the order of 1 ohm, so the time to charge your capacitor without a resistor is approximately $$ t_{real} \approx 5C $$ This resistance depends on what type of battery, how dead the battery is, etc... so this is only a rough estimate.

\$\endgroup\$
  • \$\begingroup\$ But then why we don't add the internal resistance of battery to 3 ohms ? \$\endgroup\$ – user41048 Oct 9 '15 at 1:56
  • \$\begingroup\$ The example you gave was a 3000 ohm resistor. 3000 + 1 isn't much different from 3000. \$\endgroup\$ – Greg d'Eon Oct 9 '15 at 1:57
  • \$\begingroup\$ treal≈5C so does that means it will take t=5 x 0.001 C = 0.005 seconds ? \$\endgroup\$ – user41048 Oct 9 '15 at 2:00
  • 2
    \$\begingroup\$ Also, the internal resistance of a battery is not constant, and the change is not strictly linear. It depends on its chemistry, voltage, temperature, load, etc. \$\endgroup\$ – user65586 Oct 9 '15 at 2:36
  • 1
    \$\begingroup\$ Maybe we should also add that in the real world, the capacitor isn´t perfect and will also have a resistance, ie the ESR. \$\endgroup\$ – F. Bloggs Oct 9 '15 at 11:44
2
\$\begingroup\$

In the pictured circuit, the time constant will be set by the internal resistance of the battery, the internal resistance of the capacitor, and the resistance of whatever wires connect the two. For a 9 V battery, the battery resistance is probably most important.

The time constant will indeed approach zero as those parasitics are reduced and the total resistance approaches zero.

\$\endgroup\$
0
\$\begingroup\$

The voltage current relationship in a capacitor is $$ i = c\frac{dv}{dt} $$

The voltage across capacitor can't change instantaneously as it would require infinite current according to above equation.

In an ideal case, the internal resistance of battery and resistance of connecting wires are zero. When you connect a battery directly to a capacitor without any resistance, you're asking the capacitor to change its voltage suddenly. This results in flow of infinite current (theoretically) that charges the capacitor in zero time(theoretically)

But practically the internal resistance of battery and resistance of wires can be modeled as a series resistance connected to capacitor. If this resistance is very small, this case is very close to the ideal. The instantaneous change would now cause a very large current flow and the capacitor charges very quickly. The resistance associated slows down the rate of charging as you can see from the equation:

$$ Vc(t) = V(1-e^-t/RC) $$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.