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.
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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
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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.
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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.
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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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