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Li_Kuilin_Lab1

Page history last edited by xyyue@... 5 years 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

 

Comments (1)

zahraa@... said

at 4:02 pm on Jul 6, 2015

Great Job

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