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This is a really simple question. But I'm seeing some confusing behavior in an experiment I'm trying to make sense of. So I'm trying to see if my understanding of theory is wrong.

Wikipedia says: The coulomb (unit symbol: C) is the SI derived unit of electric charge (symbol: Q or q). It is defined as the charge transported by a steady current of one ampere in one second:

Ok so 1 amp for one second transports one coulomb, right?

So in other words, if I discharge a capacitor for one second at 1 amp and charge a second capacitor for one second at 2 amps from another power source, the final state outcome is that I put twice as much charge on the second capacitor as I discharged from the first capacitor, right?

If that is correct, then let me also ask something else: The voltage is irrelevant to the measured energy expenditure right? It doesn't matter if the first capacitor starts at 100 volts and discharges to 0 volts while the second capacitor starts at 0 and charges to 10 volts. All that matters for computing the energy movement is the actual charge/amps. Right?

Also, is there any kind of known condition where this doesn't hold true or something that easily trips people up?

JamesHoux
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  • I'm not sure if it helps but capacitors store energy not electrical charge. A charged capacitor contains the same number of charge carriers as an uncharged capacitor. When you charge or discharge a capacitor, the same amount of electrical charge flows in one side as flows out the other side - net accumulation of electrical charge is zero. – RedGrittyBrick Jun 27 '13 at 09:12

5 Answers5

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If that is correct, then let me also ask something else: The voltage is irrelevant to the measured energy expenditure right? It doesn't matter if the first capacitor starts at 100 volts and discharges to 0 volts while the second capacitor starts at 0 and charges to 10 volts. All that matters for computing the energy movement is the actual charge/amps. Right?

This part is not right. The voltage is very relevant to the energy.

The energy stored in a capacitor is:

$$E = \frac{1}{2}CV^2$$

The voltage on a capacitor is determined by the charge and the capacitance:

$$V = \frac{Q}{C}$$

If you have a smaller capacitor, a given amount of charge will create a higher voltage, and it will take significantly more energy to put it there.

Dave Tweed
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The problem (or rather, what may make this confusing) is that your thought experiment is not actually that trivial to turn into a real experiment. The simple capacitor experiments involve hooking up the cap to a near constant voltage source, which will give you the well-known charging graphs (charging current starts out high, then decreases).

Anyways, let's say you do have a constant current source to charge your caps with. If you charge C1 at 1A for 1s, you put 1C of charge on it. If you charge C2 at 2A for 1s, you put 2C of charge on it.

If you then disconnect the current source, the voltage between the two leads will be determined by the capacitance: If C1 ends up at 100V, that means it has a capacitance of 1C/100V = 10mF. If C2 ends up at 10V, it has a capacitance of 2C/10V = 200mF.

(For an explanation of the energy issues involved, see Dave Tweed's answer.)

us2012
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One ampere is one coulomb per second, yes. Therefore, one coulomb is one amp-second.

The official definition of the farad is the amount of capacitance on which a charge of one coulomb will change its voltage by 1 volt.

We know that:

\$I_C = C \dfrac{dV}{dt} \$

or

\$ dV = \dfrac{I_c \cdot dt}{C}\$

Therefore:

\$ V = \dfrac{I_c \cdot t}{C} = \dfrac{Q}{C}\$ (omit the deltas for now)

\$ C = \dfrac{Q}{V}\$

Charge is charge regardless of the voltage. 1 coulomb is 1 coulomb. The capacitor voltages will vary depending on their sizes (the larger the cap, the lower the voltage a coulomb will produce).

Dave Tweed
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Adam Lawrence
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Your understanding about the meaning of farad is correct.

Your reasoning about the energy expenditure is not. Here's one way to understand the wrongness: if power is the rate of energy use or conversion, and only current (and not voltage) was relevant for calculating energy, then why would electric utility companies transmit electrical energy over long distances at high voltage? If you know that \$P=IE\$ then you know that voltage is very relevant to electrical energy in general, capacitors being no exception.

Or, understand that voltage is a difference in potential of an electric field. Difference in potential in a gravitational field can be expressed by height. Gravity fields exert forces on things with mass. Electric fields exert forces on things that have charge. Does the height of a mountain have any relevance on the energy expenditure required to push a mass from the base to the top, or is just the mass relevant?

You might also want to read Can a charge-pump be 100% efficient, given ideal components?

Community
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Phil Frost
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All that matters for computing the energy movement is the actual charge/amps. Right?

This is most certainly wrong. The voltage must be relevant.

In fact, the change of energy associated with a capacitor is expressed in terms of the voltage across the capacitor, not the charge.

First, the energy stored by a capacitor is quadratic in the voltage:

\$E = \dfrac{1}{2}Cv^2\$

Thus, the change in energy is given by:

\$\Delta E = \dfrac{1}{2}C(v_F^2 - v_I^2)\$

If you wish to know the power associated with a capacitor (or any other circuit element), simply form the product of the voltage and current associated with the capacitor:

\$p_C = v_C \cdot i_C = C v_c \dfrac{dv_c}{dt}\$

Which is, again, in terms of the voltage across the capacitor.

Alfred Centauri
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