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In a circuit there are two capacitors, Capacitor a and Capacitor b, now both of them have the same voltage values but different farad values, now, I know this must be a basic question, but I'll like to know: What is the difference does the capacitors make in the circuit, they both have the same voltage value, is it important to prioritize the farads value as well?

Basically, am asking: what is the unique difference between a capacitor with maybe like 220uf 12v and another capacitor with 2200uf 12v?

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    \$\begingroup\$ The 2200 uF capacitor has a 10x larger value. What the influence of the value of a capacitor is in a circuit depends on the circuit. You do not show a circuit so the only answer that can be given is the first sentence of this comment. \$\endgroup\$ Commented Jun 30, 2019 at 17:40
  • \$\begingroup\$ dt/dt=Ic/V. .... \$\endgroup\$ Commented Jul 1, 2019 at 0:05
  • \$\begingroup\$ It is a very basic question you need to do some research on what capacitance is. \$\endgroup\$
    – RoyC
    Commented Jul 1, 2019 at 1:59
  • \$\begingroup\$ tinyurl.com/yy8vob3b play here \$\endgroup\$ Commented Jul 1, 2019 at 4:53

2 Answers 2

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The charge stored in a capacitor is given by \$ Q = CV \$, where Q is the charge (C, coulomb), C is the capacitance (F, farads) and V is the voltage (V, volt).

The 2200 μF capacitor holds ten times the charge of the 220 μF capacitor when both are at the same potential.

enter image description here

Figure 1. Ripple voltage from a full-wave rectifier, before and after the application of a smoothing capacitor. Source: Wikipedia - ripple.

If, for example, these were used on a rectifier circuit to smooth out mains voltage ripple then the 2200 μF would do a ten times better job than the 220 μF.

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  • \$\begingroup\$ Thank you! Exactly the answer that I was looking for!! \$\endgroup\$ Commented Jun 30, 2019 at 17:50
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    \$\begingroup\$ Glad to help. Wait a day or two to encourage some more answers which may give you other insights. Then accept the most helpful and upvote any other good ones. \$\endgroup\$
    – Transistor
    Commented Jun 30, 2019 at 18:06
  • \$\begingroup\$ 👍you've got it. \$\endgroup\$ Commented Jun 30, 2019 at 18:57
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I like Transistor's answer, here's another view of it if you want to take a stab at it in a energy or power direction. Maybe it's too much, oh well.


The energy stored in a capacitor is \$\dfrac{CV^2}{2}=C\dfrac{V^2}{2}\$ joules.

  • \$(220\text{µF})\dfrac{(12\text{V})^2}{2}\approx 16 \text{mJ}\$
  • \$(220\text{µF})\dfrac{(38\text{V})^2}{2}\approx 160 \text{mJ}\$
  • \$(2200\text{µF})\dfrac{(12\text{V})^2}{2}\approx 160 \text{mJ}\$

Given the same voltage, you can say that one has 10x more energy than the other. If the 220µF one were to have the same energy has the 2200µF one, then the voltage of the 220µF one would have to be about \$12\sqrt{10}\approx 38\text{V}\$.

Another way of looking at it is that there is one factor that determines a capacitors capacitance that looks like this: \$\frac{\text{overlapping area of plates}}{\text{distance between plates}}=\frac{a}{d}\$. So maybe the other one has an area that is 10x larger than the other one and same distance between the plates. Or maybe it has the same area but has a distance that is a 10th of the other one between its plates. Or a mix of the two, whatever that makes \$\frac{a}{d}=10\$


Tiny example to put numbers to it:
Let's say that you have some kind of voltage regulator (step down, 100% efficient) that gives 5V out and takes 5V to 12V as input. And you want to power an Arduino microcontroller (µC) that consumes 10mA @ 5V for 7 seconds. Then how much capacitance must your capacitor have at least?

The energy consumed by a component is power × time, power is ampere × voltage.

  • µC needs \$(10\text{mA})(5\text{V})(7\text{s})=0.35\text{mJ}\$

The usable energy in the capacitor (whose capacitance we need to determine) becomes \$C\dfrac{V_{high}^2-V_{low}^2}{2}\$

This becomes \$C\dfrac{12\text{V}^2-5\text{V}^2}{2}=0.35\text{mJ}\$

Move things around and you get \$C=\dfrac{2(0.35\text{mJ})}{12\text{V}^2-5\text{V}^2}\approx 5.9\text{µF}\$

So if you want to power a µC that consumes 10mA @ 5V for 7 seconds and you have a 100% efficient voltage regulator and your source is a capacitor whose voltage will start at 12V and go to 5V as it powers the µC. Then the capacitance of that capacitor must be at least 5.9µF.

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