1.Does the current flow through the capacitor when (a) it's not charged, (b) it's partially charged, (c) it's fully charged ? How does the flow changes in time (ex.: exponentially, logarithmically, linearly, etc.)?

2.If you charge a 16V capacitor from a 9V battery and then you disconnect the battery, what will be the capacitor voltage? 16 or 9?

  • 2
    \$\begingroup\$ Is this homework? \$\endgroup\$
    – starblue
    Commented Apr 28, 2012 at 11:55

3 Answers 3


If it's fully charged there's no current. When not or partly charged there will be current if you apply a voltage to it. The current curve depends on how your voltage source is connected. If you supply a constant current the voltage will rise linearly, if you charge over a resistor it will rise exponentially. That's because the rising voltage will cause the voltage drop over the resistor to become smaller and smaller, and hence the current through it will also decrease. A lower current means that charging goes slower.

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How fast the capacitor charges depends on both the resistor's and the capacitor's value. The larger the resistor the slower the capacitor charges, and it also charges slower if the capacitance is greater. The exponential curve is said to have a time constant \$RC\$. After about \$5 \times RC\$ the capacitor is charged to 99%.

9V. The capacitor's rating has nothing to do with it, that only determines how high a voltage you can apply. Just look at what voltage is available, and that's 9V. Just take care that the capacitor is fit for the supplied voltage. You can safely charge a 16V capacitor to 9V, but not to 25V.

  • \$\begingroup\$ yes son , there's no easy way to learn this without learning differential equations. \$\endgroup\$ Commented Apr 27, 2012 at 13:34
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    \$\begingroup\$ No DC current ideally... :P \$\endgroup\$
    – Nick T
    Commented Apr 27, 2012 at 20:09

You seem to have some basic misconceptions about capacitors. The current thru a capacitor is proportional to the change in the voltage. In more mathematical terms:

 current = K dV/dt

The proportionality constant K is the capacitance. When the units are Amperes, Volts, Farads, and seconds, it all works out with a scale factor of 1:

 Amps = Farads * Volts / second

So if nothing is externally forcing the voltage of a capacitor to change, there will be no current thru it. Equivalently, if nothing is forcing current thru a capacitor, it's voltage won't change. If you charge up a capacitor to 9 V by connecting it accross a 9 V battery and waiting long enough for the current to go to zero (the voltage to stop changing), then disconnect it, the voltage on the capacitor will stay 9 V. That is for a perfect capacitor. Real capacitors have leakage, meaning a little current will leak thru them to discharge them, as if a large resistor was connected accross them.

The voltage rating of a capacitor only tells you how many volts you can charge it up to before it may break. So charging up a 16 V capacitor to 9 V is perfectly fine since 9 V is less than 16 V, but that 16 V rating has otherwise nothing to do with how many volts will be on the capacitor in a circuit.


Keep in mind many caps have self leakage , so may decay fast.

Plastic types (polypropylene) are much better than electrolytics which are better than most ceramics.

If you connect a discharged cap to a charged Cap. YOu can predict the voltage change from the cap value or visa versa. since total charge remains almost same. Q=CV

  • \$\begingroup\$ While you're right, I think it's a bit too high level for the kind of question \$\endgroup\$
    – clabacchio
    Commented Apr 27, 2012 at 19:11

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