# Difference between revisions of "Circuit Lab/Episodes"

(First bunch of things from the old wiki. I'm moving the rest over in a few minutes) |
(→Circuit Lab: Part 2 Posting) |
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(that last one is more advanced) | (that last one is more advanced) | ||

+ | |||

+ | ===Episode II=== | ||

+ | |||

+ | A little review, perhaps: | ||

+ | |||

+ | The three concepts of Voltage, resistance, and current are all interrelated through this basic formula: | ||

+ | |||

+ | V = I R | ||

+ | |||

+ | or Voltage = Current times Resistance | ||

+ | |||

+ | |||

+ | '''Okay, whatever Demosthenes, I don't really care, why do I have to learn this formula anyways? Like I care about this 'omg' guy or whoever he is.''' | ||

+ | |||

+ | You have to learn Ohm's law because it helps you to 'analyze' circuits. That means you can use this law to find voltage, resistance, or current, if you have two of the three. Let's look at how to apply this formula: | ||

+ | |||

+ | The application of this formula is pretty easy, once you get the hang of it. Basically, imagine a wire, a battery, and a resistor somewhere along the wire. If the battery has a voltage of 10 volts, and the resistor has a resistor of 30 ohms, you simply use Ohm's law to find the current: | ||

+ | |||

+ | V= IR .....write your equation, so you know what you're doing here... | ||

+ | |||

+ | 10 volts = (I)(30 ohms) ...Set up the equation, plugging in the values.... | ||

+ | |||

+ | (10 volts)/(30 ohms) = I ...Divide both sides by '30 ohms' so that you can isolate the variable, I, or the current.... | ||

+ | |||

+ | 10 volts/30 ohms = 1/3 amperes do out the math....fun!... | ||

+ | |||

+ | '''But wait Demosthenes, what if they ask for voltage or resistance?''' | ||

+ | |||

+ | Don't get scared, my young padawan. The equation can be set up so that no matter which two of the three variables you know, you can figure out the other one easily. Suppose there's a circuit with a 6 volt battery and 2 amps of current, how would you set that up? What's your answer? (You try it first, and see if it agrees with mine!!) | ||

+ | |||

+ | Alright, let's see how you did: | ||

+ | |||

+ | V = IR ....okay, first write out the equation so you know what you're doing | ||

+ | |||

+ | R = (V)/(I) .....Manipulate the equation so you have the two knowns on one side | ||

+ | |||

+ | R = 6 volts / 2 amps ....Plug in the values | ||

+ | |||

+ | R = 3 ohms ....solve by dividing 6 by 2. | ||

+ | |||

+ | Here are the 3 general forms of the law you'll need to know: | ||

+ | |||

+ | V = IR R = V/I I = V/R | ||

+ | |||

+ | Whichever value you're searching for, simply make that the 'lone' variable, then plug in the values, and see what you get. Pretty simple. | ||

+ | |||

+ | There's some nice little quiz questions at the bottom of the page in the following link which you can test yourself further with... | ||

+ | |||

+ | [http://www.seattlerobotics.org/guide/electronics.html Go to the bottom...there's questions and answers there!] | ||

+ | |||

+ | [http://library.thinkquest.org/10784/circuit_symbols.html A Guide to Electric Symbols] |

## Revision as of 17:20, 11 July 2008

# Circuit Lab

### Episode I

-in-the-fingers-

**-What is a 'circuit'?**

- ) end. When you touch a wire onto both ends of the battery at the same time, you have created a circuit. What just happened? Current flowed from one end of the battery to the other through your wire. Therefore, our definition of circuit can simply be a never-ending looped pathway for electrons (the battery counts as a pathway!).

**-**

**-Oh, so I get it Demosthenes, you could just put a wire onto one end of a battery, and the electrons would still bump each other?**

No, you could not. As stated before, in our definition of the circuit, a continuous loop is required. But think about it scientifically: If you did attach the wire to only one end of the battery, where would the electrons go that got bumped to the opposite end of the wire? That is why there needs to be that continuous loop of wire: the electrons need somewhere to go.

**-**
-so-

Or

Voltage = Current times Resistance

If you are having trouble, think back to the baseball example: you can have a high chance of winning (voltage) by either scoring a lot of runs (high current) or having good defense/pitching (resistance).

**-**

Have you ever pumped up a super----

**-**

(that last one is more advanced)

### Episode II

A little review, perhaps:

The three concepts of Voltage, resistance, and current are all interrelated through this basic formula:

V = I R

or Voltage = Current times Resistance

**Okay, whatever Demosthenes, I don't really care, why do I have to learn this formula anyways? Like I care about this 'omg' guy or whoever he is.**

You have to learn Ohm's law because it helps you to 'analyze' circuits. That means you can use this law to find voltage, resistance, or current, if you have two of the three. Let's look at how to apply this formula:

The application of this formula is pretty easy, once you get the hang of it. Basically, imagine a wire, a battery, and a resistor somewhere along the wire. If the battery has a voltage of 10 volts, and the resistor has a resistor of 30 ohms, you simply use Ohm's law to find the current:

V= IR .....write your equation, so you know what you're doing here...

10 volts = (I)(30 ohms) ...Set up the equation, plugging in the values....

(10 volts)/(30 ohms) = I ...Divide both sides by '30 ohms' so that you can isolate the variable, I, or the current....

10 volts/30 ohms = 1/3 amperes do out the math....fun!...

**But wait Demosthenes, what if they ask for voltage or resistance?**

Don't get scared, my young padawan. The equation can be set up so that no matter which two of the three variables you know, you can figure out the other one easily. Suppose there's a circuit with a 6 volt battery and 2 amps of current, how would you set that up? What's your answer? (You try it first, and see if it agrees with mine!!)

Alright, let's see how you did:

V = IR ....okay, first write out the equation so you know what you're doing

R = (V)/(I) .....Manipulate the equation so you have the two knowns on one side

R = 6 volts / 2 amps ....Plug in the values

R = 3 ohms ....solve by dividing 6 by 2.

Here are the 3 general forms of the law you'll need to know:

V = IR R = V/I I = V/R

Whichever value you're searching for, simply make that the 'lone' variable, then plug in the values, and see what you get. Pretty simple.

There's some nice little quiz questions at the bottom of the page in the following link which you can test yourself further with...