03_handson_activity_discharging_Capacitor

This is part of my hands-on activity - 01 Course : Computer Engineering - UNINTER - Brazil - Electric Circuit Analysis II Date: march, 2022 Author: Gilberto Oliveira Junior RU 3326662 Materials: We will use: DC = 12v C = 1000uF R = 3.3k ohms How to run the experiment: To make it run both activities, run normally, as usual. Click o the graphics and you are good to go! A capacitor (1000 uF) is charged by connecting it to a DC voltage source using a resistor R = 3300 Ohms. Our DC power supply is rated to 12 V. Once the capacitor is connected to the DC voltage source, it will charge up to the voltage that the DC voltage source is outputting. The capacitor takes 5-time constants to charge to its maximum rating. The time constant, τ, is found using the formula T = R*C in seconds. Therefore the time constant τ is given as: 5*τ= 5 * (3.300) * (1000 * 10^-6) = 16.5 seconds Thank you. Bye o/