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2. 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.

To get 25 mA of current using a green LED, simply apply Ohm’s law(V=IR). To include the voltage drop in the LED, subtract the maximum voltage rating for 25mA (2.5V) from the initial 5V produced in the power supply. In this case current, I, would be equal to 25e-3 A and the voltage, V, would be (5-2.5)V. Thus, 2.5V/25e-3V = 100.

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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 determined above is a minimum value for resistance because using any less resistance with a constant voltage drop, demands a higher current, and this will damage the component. This behavior is due to the inverse relationship between current and resistance as seen in Ohm’s law. In other words, a lower resistance will make the LED prone to fail.

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c. What is the resistance range of the potentiometer?

The resistance range of the potentiometer is between 5 and 18 with 5 being the minimum and 18 the maximum ¾ of a turn away.

It should be about 10K

But I won’t deduct any points.

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

The order of the circuit components does not matter as long it is all arranged in series but once there are parallel connections involved it can matter because the current splits between the leads. But within a series connection the order does not affect anything besides possibly functionality because the voltage over each component will be the same and the current will be

Great Answer

constant.

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

Similarly to the initial equation to determine the resistance required, this situation calls for Ohm’s law(V=IR). Accounting for the voltage drop of the green LED, the voltage going through the necessary resistor if 9V - 2.5V -> 6.5V. Then apply this to Ohm’s law using the max current(25mA). 6.5V = (25e-3 A)R Solving would result in a minimum R of 260.

## Comments (1)

## zahraa@... said

at 2:31 pm on Jul 6, 2015

Good Job

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