| 
  • If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

  • Stop wasting time looking for files and revisions. Connect your Gmail, DriveDropbox, and Slack accounts and in less than 2 minutes, Dokkio will automatically organize all your file attachments. Learn more and claim your free account.

View
 

Lab1 (1)

Page history last edited by Ajay Sohmshetty 7 years, 5 months ago

Ajay Sohmshetty

Question 2

  1. I need a resistance of 105 Ohms to limit the current to 30mA using the red LED. Our voltage is 5 – 1.85V. So, V = 3.15V. I = 0.03A. Thus, R = V/I, so R = 3.15/0.03 = 105 Ohms.

  2. The resistance from question a) is a minimum resistance. That is, if you decrease the resistance, the LED will likely fail. If you lower the resistance, then by V = IR, since V is constant, I will increase, and it will increase above the 30mA. Thus, the LED will fail if you lower the resistance.

  3. The resistance range of the potentiometer is from 1Ohm to around 9k-10k Ohms. I did this by detecting the voltage drop and used V = IR to calculate the resistance.

Question 3

  1. It not does matter what order the components of my circuit are arranged between power and ground. Since the components are in a series circuit, and not a parallel circuit, if the circuit is complete then current will flow through all of the components at a constant value. Thus, each component will have the same R, V, and I values.

Question 4

  1. Using this 9V battery, the minimum resistance required for use with my LED is 300Ohms. V =9V, I = 0.03A. R = V/I. So, R = 300 Ohms.

Video:

http://www.youtube.com/watch?v=KCmpNbp6OBo

 

Comments (0)

You don't have permission to comment on this page.