Lab 1: Orientation and LED
2.
a)
Voltage draw of green LED (the LED I used) = 2.2 V
Voltage of source = 5V
25mA = .025A
5-2.2=.025*R
2.8/.025 = R = 112 Ohms.
b)
The resistance from question 1a is a minimum resistance. I.e. by bringing the resistance any lower than this the LED would be likely to fail.
c)
The range of resistance of the potentiometer is .241 Ohms to 9.97 KOhms.
3.
a)
It does not matter in which order the components of my circuit are arranged between power and ground because there is only one path for the electric current to follow, therefore regardless of where the switch is when it is closed the current will have a path to follow and the LED will light and when it is open there will be no complete path to follow so the current will not flow and the LED will not light.
4.
a)
The minimum resistance required for use with my LED using a 9V battery is:
let R be the minimum resistance.
9V-2.2V = .025*R
6.8/.025 = R = 272 Ohms
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