a. What resistance do you need to limit current to 30 mA (if using red LED) or 25 mA (if using yellow or green)? Be sure to state which color LED you are using. This resistance refers to the total resistance in series with the LED.

According to Ohm’s law, V = IR, or Voltage = Current x Resistance. We know that our initial voltage is 5 Volts, and the Red LED light causes a 1.8 volt drop in the circuit, so our total voltage would be 3.2 Volts. If we plug in the Voltage and the .03 Amps into Ohms law, we get 3.2V = .03A * ?Ohms, and if we solve for ?Ohms, we get ? = 107 Ohms.

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b. Is the resistance from question a) a maximum or minimum resistance? That is, in which direction if you change the resistance (higher or lower) would the LED likely fail.

The resistance is a minimum, and a lower resistance would cause the LED to fail.

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c. What is the resistance range of the potentiometer?

The potentiometer ranges from 0 to 10000 Ohms

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a. Does it matter what order the components of your circuit are arranged between power and ground? Why or why not?

No, it doesn't matter what order the components of the circuit are arranged, so the resistor and LED could be arranged in any way, as long as it is a complete circuit.

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a. Using this battery, what is the minimum resistance required for use with your LED?

Using the 9V battery, we basically repeat the process we did in the first question. Since we know the voltage drop for the red light is 1.8V, we get 7.2 Volts. Using Ohms law, we plug in 7.2 for Volts, 30mA for Current, and ?Ohms. So we get 7.2V = .03A *?Ohms. When we solve for ?, we get 240 Ohms.

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## Comments (1)

## Vivien Tsao said

at 6:17 pm on Jul 11, 2013

Good job! :)

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