If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.
You already know Dokkio is an AI-powered assistant to organize & manage your digital files & messages. Very soon, Dokkio will support Outlook as well as One Drive. Check it out today!

Li_Kuilin_Lab1

Page history last edited by xyyue@... 8 years, 9 months ago

a. The forward voltage drop of the LED is 2.5 volts and it runs at 25 mA. At 5 volts, to limit the circuit to 25 mA = 0.025 A, with Ohm’s Law with respect to the resistor, its voltage drop should be 5-2.5 = 2.5 V, and so its resistance should be 2.5/0.025 = 100 Ohms.

✔

b. 100 Ohms is the minimum resistance since it is on an inverse relationship with the current in the circuit, which is limited at 25 mA

✔

c. Measured by the multimeter, 0 to 2.5KOhms.

It should be 10K Ohms, -0.5

a. No, because the resistor will resist current wherever it’s placed so long as it is in series, and the switch also can break the circuit at anywhere in series with it.

✔

a. Using a 9V battery, the voltage drop across the LED should still be 2.5 volts, and so the voltage drop across the resistor should be 9-2.5 = 6.5 volts, and when the current is 25 mA = 0.025 A, its resistance should be 6.5/0.025 = 325 ohms minimum.

✔

Video: http://www.youtube.com/watch?v=JaSqckXZub8