Capacitance
Q = C · U
| Quantity | Symbol | Unit |
|---|---|---|
| Charge | Q | C |
| Capacitance | C | F |
| Voltage | U | V |
Elec Class
learn electricity hands-on
Lesson · chapter 9 of 15
A capacitor stores electric charge: with a voltage U across it, it builds up a charge Q = C × U, where C is its capacitance in farads. Charged through a resistor R, it is full after about 5 × R × C seconds.
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used Q = C · U
3The result
Charge Q5.64 mC
A 470 µF capacitor charged to 12 V stores a charge of 5.64 mC.
The definition, the key points to remember and everyday examples.
A capacitor is made of two conducting plates separated by an insulator. At a voltage U it stores a charge Q = C·U; C is its capacitance, in farads (in practice µF, nF, pF). It stores the energy W = ½ C U².
In parallel capacitances add up; in series their inverses add up. Charged through a resistor, it reaches 63 % of its final voltage after τ = R·C and ≈ 100 % after 5τ; in steady DC it passes no current.
A capacitor is like a tank filled through a pipe. The wider the tank (large capacitance C), the more water it holds for the same height (the voltage U): that's Q = C × U. A thin pipe (large resistance R) slows the filling: that's the time constant τ = R × C. Where it falls short: no charge crosses the insulator; whatever arrives on one plate leaves from the other, and the capacitor stays neutral overall.
| Quantity | Symbol | Unit |
|---|---|---|
| Charge | Q | C |
| Capacitance | C | F |
| Voltage | U | V |
| Quantity | Symbol | Unit |
|---|---|---|
| Energy | W | J |
| Quantity | Symbol | Unit |
|---|---|---|
| Equivalent capacitance | Céq | F |
| Quantity | Symbol | Unit |
|---|---|---|
| Time constant | τ | s |
| Resistance | R | Ω |
| Capacitance | C | F |
3-digit marking: two digits then number of zeros, in pF (104 = 100 000 pF = 100 nF).
Four steps to solve a problem, then an example solved line by line.
µF = 10⁻⁶ F, nF = 10⁻⁹ F, pF = 10⁻¹² F; and kΩ to Ω for the time constant.
Replace each group with its equivalent capacitance: in parallel you add them, in series you add the reciprocals.
Q = C × U for the charge, W = ½ × C × U² for the energy, τ = R × C for the charging time.
In electronics, Q is counted in µC or mC and τ in ms or s: a result in kilocoulombs gives away a forgotten power of 10.
A 100 µF capacitor is charged at 12 V. What charge and energy does it store? It is charged through R = 10 kΩ: time constant and charging time? Two such capacitors in series: equivalent capacitance?
Q = C · U = 100 × 10⁻⁶ × 12 = 1.2 × 10⁻³ C = 1.2 mC.
W = ½ · C · U² = 0.5 × 100 × 10⁻⁶ × 144 = 7.2 mJ.
τ = R · C = 10⁴ × 10⁻⁴ = 1 s: 63 % of 12 V (7.6 V) after 1 s, full charge (99 %) after 5τ = 5 s.
In series: 1/Ceq = 1/100 + 1/100, Ceq = 50 µF.
Multiply its capacitance by the voltage: Q = C × U. A 470 µF capacitor at 12 V stores Q = 470 × 10⁻⁶ × 12 ≈ 5.6 × 10⁻³ C, or 5.6 mC.
It's the capacitance of a capacitor that stores 1 coulomb at 1 volt. It's a huge unit: common capacitors range from a picofarad (10⁻¹² F) to a few thousand microfarads; only supercapacitors reach several farads.
The voltage (for example 25 V) is the maximum not to exceed. The stripe marks the − lead: an electrolytic capacitor is polarized, and connected the wrong way round it heats up and can swell or burst.
About 5 × R × C through a resistor R: with each time constant, its voltage is multiplied by about 0.37. A 1,000 µF capacitor through 10 kΩ has τ = 10 s: it is practically empty after 50 s.
Because it keeps its charge until the charge finds a path to flow away. Those in a power supply or a microwave oven can stay charged to a dangerous voltage after the power is cut: they are discharged through a resistor before any work is done.
A capacitor stores energy in electrical form and can give it back in a fraction of a second; a battery stores it in chemical form, in far greater amounts. 1,000 µF at 12 V holds 0.072 J, an AA battery about 10,000 J.
The voltage rises as u = U (1 − e^(−t/τ)): at t = τ, 1 − 1/e ≈ 0.63. At 3τ it is 95 %, at 5τ more than 99 %: the charge is considered complete.
To store energy and release it (flash, buffer), to smooth a rectified voltage, to time a delay (RC circuit). In steady state it draws nothing.
From the simplest to the most complete, each problem comes with a step-by-step solution.
Charge a 1,000 µF capacitor from 9 V through 10 kΩ (τ = 10 s), record its voltage with the voltmeter every 5 seconds and plot the curve: you read about 5.7 V (63 %) after 10 s.
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