How to Charge a Single Capacitor in a Series With a Resistor

Like a battery, a capacitor is an electronic device designed to store electrical charge. Unlike a battery, a capacitor charges and discharges at a very rapid rate that is practically instantaneous. To slow down the rate of charge and discharge, you can place a resistor in a series with a capacitor. In addition, an LED (light emitting diode) is also typically included in the series circuit so that you can visually monitor the charge cycle.

Things You'll Need

  • 25-volt electrolytic capacitor
  • Variable DC power supply
  • Electronic breadboard
  • Hookup (jumper) wire
  • Red LED
  • 1000-ohm resistor
Show More

Instructions

    • 1

      Connect the negative lead of a variable DC power supply to the negative bus in an electronic breadboard. Connect the positive lead of the power supply to the positive bus in the breadboard.

    • 2

      Insert the leads of a 1000-ohm resistor into one section of the breadboard. Next, insert the leads of a red LED into a separate section of it. Insert the leads of a 25-volt capacitor into yet another separate section of the breadboard. Breadboards consist of multiple sections electrically isolated from each other so that you can insert the components separately. You can then use short lengths of hookup (jumper) wire to "jump" from one section to another to connect the components.

    • 3

      Connect a jumper wire from the positive bus to the positive lead of the red LED. The longer of the two leads is the positive (anode) lead. Connect a jumper wire from the remaining LED lead to either of the two leads of the resistor.

    • 4

      Connect a jumper wire from the remaining resistor lead to the positive lead (anode) of the capacitor. Next, connect a jumper wire from the remaining lead (cathode) of the capacitor to the negative bus lead in the breadboard.

    • 5

      Turn on the power supply and adjust the voltage control knob to 25 volts. Observe the red LED. While the red LED is alight, current is flowing and the capacitor is charging. When the LED goes out, current is no longer flowing and the capacitor is fully charged.

Learnify Hub © www.0685.com All Rights Reserved