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# COsorio Lab1

last edited by 8 years, 2 months ago

Lab 1
Part C
2a. 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.
For a red LED, the voltage drop is 1.85V. This means that 5V-1.85V=
3.15V will drop across the resistor. Using a 5V supply, a resistance
of
R=V/I=3.15V/.03A=105 ohms
is needed to limit the current to 30 mA.
For a green LED, the voltage drop is 2.2V. The voltage drop across the
resistor will be 5V-2.2V=2.8V. To limit the current to 25mA, the
resistance should be at least
R=V/I=2.8V/.025A=112 ohms
For a yellow LED, the voltage drop is 2V. The voltage drop across the
resistor will be 5V-2V=3V. To limit the current to 25mA, the
resistance should be at least
R=V/I=3V/.025A=120 ohms

2b. 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 resistance. By increasing the resistance,
we can further decrease the current. This means that it is safe to
increase the resistance, while decreasing the resistance will increase
the current and may lead to an LED failure.

c.What is the resistance range of the potentiometer?
.3 ohms - 9.76 kohms.

3a. Does it matter what order the components of your circuit are
arranged between power and ground? Why or why not?
The order that the components are arranged between power and ground is
not important. Because all of the elements are in series, the current
through all of the elements is the same.

4a. Using this battery, what is the minimum resistance required for
For the red LED, the minimum resistance will be
R=V/I=(9V-1.85V)/.030A=238.34 ohms.
For the yellow LED
R=V/I=(9V-2V)/.025A=280 ohms.
For the green LED
R=V/I=(9V-2.2V)/.025A=272 ohms. 