Physical Computing Homework 2

Homework assignment 2

Part 1
https://drive.google.com/file/d/0B8ZGdT5f3eJkczBsSkswcXNHUkk/view?usp=sharing

Part 2
https://drive.google.com/file/d/0B8ZGdT5f3eJkR1hzeFpNWFVNN28/view?usp=sharing
1) I would move the switch past the junction point along the 560 ohm resistor.
Wo when the switch is open, current can only flow through the 10kiloOhm resistor.
When the switch is closed, most of the current will flow through the 560 ohm resistor due to the significantly lower resistance.
Arguably no current goes through the 10kiloOhm resistor.

Part 3
https://drive.google.com/file/d/0B8ZGdT5f3eJkTklKbS1yX0QtUlk/view?usp=sharing
1) 5 Volts, effectively no resistance means all of the voltage is left going back into the pin.
2) 0 Volts, all voltage is used up in the resistor.
3) 5kOhms because this evens out the current going into both sides, and since 10kOhms takes up all the voltage, 5kOhms will take up half as much voltage.
4) Because the LED faces the full force of voltage coming from pin5 since the resistor is located after.
In this situation, 5V / 4 results in a maximum voltage of 1.2V that will protect the LED from blowing.

Part 4
https://drive.google.com/file/d/0B8ZGdT5f3eJkQ3o2cmdPMk1PbEE/view?usp=sharing
1) No, becuase the speaker has 8Ohms of resistance which limits the current and prevents a short circuit.
2) More current.
3) Yes.

Part 5
https://drive.google.com/file/d/0B8ZGdT5f3eJkd3RlQ3ZKUzBFc3M/view?usp=sharing
1) Switch the locations of the 10kOhm resistor and photoresistor. When light increases, resistance increases on the path going to ground, which means more current flows into the pin.

Instrument:

By the definition of interactivity, there is a cycle of exchange between me, the controller and the unstruments base note. As the instrument continues to play arpeggios, I adjust my input so that it adjusts it's output.

Critique: No cover. could be better on that aspect.

https://drive.google.com/file/d/0B8ZGdT5f3eJkNjdmaFVtR1E3OEE/view?usp=sharing

Week 1 Physical Computing: Basics
by Simon Zhang

This week we worked with basic electronics using resistors, capacitors, potentiometers, LED's and switches.

Simple LED Circuit
https://drive.google.com/open?id=0B8ZGdT5f3eJkNDZsc2I2ZWVHdnc&authuser=1

1) What is the purpose of the resistor?

Protect the LED from being fried. To use up some of the voltage and to limit current so that wires do not overheat.

2) Using Ohm’s Law (V=IR), what is the current running through this circuit?

5/560 = .009 Amps3) Using Ohm’s Law again, calculate the resistor value you would have to use in order to have only 15mA of current in this circuit?

5/.015 = 333.33333~ Ohms

4) Assuming the LED drops 2.2V of voltage, how much voltage is dropped by the resistor?

5-2.2 = 2.8Volts

LED Circuit with Switch

https://drive.google.com/open?id=0B8ZGdT5f3eJkSEhJUUpEd28yQ2M&authuser=1

1) Looking at the schematic diagram, if you moved the switch so that it was connected between the resistor and the LED, how would the circuit’s behavior change?

Nothing would Change. The switch in any location on this circuit acts to open and close the circuit.

2) Looking at the schematic diagram, if you moved the switch so that it was connected between the LED and ground, , how would the circuit’s behavior change?

Still Nothing would change. For the same reason as before.

Simple LED Circuit with Potentiometer

https://drive.google.com/open?id=0B8ZGdT5f3eJkWi1NYW5wQ1NHbkU&authuser=1

1) Why is the 560Ω resistor necessary?

Once again, this is to limit the current and reduce the maximum voltage allowed to go into the LED.

If the Potentiometer were turned so that resistance towards the LED were 0, then we depend on the 560 Ohm resistor to keep the LED from frying.

2) How much current is flowing through the circuit when the 10kΩ potentiometer is turned all the way up (offering the maximum resistance)?

5/(10000+560) = .5 milliAmperes

This effectively makes the LED go out since there is too much resistance.

3) What other types of variable resistors do you know about (name at least 4)?

Volume control, Brightness control, Photoresistors.

Dueling LEDs circuit with potentiometer

https://drive.google.com/open?id=0B8ZGdT5f3eJkQ2VGeEc1ZVNKaFE&authuser=1

1) Give a quick (1-2 sentences) explanation of why this circuit behaves the way it does. Include an explanation for how the current going through each LED is affected by a single turn of the potentiometer

The potentiometer just varies the resistance going towards both sides of the output wires. By turning it all the way one way, one directions recieves all 10kOhms of resistance and the other side has no resistance. By turning the potentiomter back and forth, the current going to both LED's varies because their currents vary. If turned all the way one way, one of the LED's would receive all remaining voltage and the other LED would have no current going through it because resistance is too high.

Capacitor Charging

https://drive.google.com/open?id=0B8ZGdT5f3eJkYUNzRURjX243VUU&authuser=1

1) How does adding a 2.2kΩ resistor between the pushbutton and the capacitor affect the capacitor’s charging behavior?

The charging behavior would be slower, and the initial current with which the capacitor is charged would start at a lower value.

Capacitor Discharging with LED delay

https://drive.google.com/open?id=0B8ZGdT5f3eJkTDZDZ0hSTFd0QWc&authuser=1

1) Looking at the circuit schematic diagram, why does the capacitor discharge up through the LED part of the circuit, and not directly from the capacitor down to ground?

Charge is built up on one side of the capacitor and can only leave through that side. In our theory, positive charge builds up on the north side of our capacitor and can only escape by moving up, turning right, and going through the LED. That way, it can travel to ground.

Test Post

Hi.