fridays post

Hey sorry for scribing late but this was the first time I was able to this weekend. All we did in class was the practise sheet ,Compound circuits, and we all teamed up in did an extra sheet, power in electric circuits, on the whiteboards. So here's the work.




We also did that extra sheet but unfortunately I'm not able to post all that. But besides the point here is what I did, pretty sure its correct but might want to make sure monday. I took total current going into the parallel circuit thats part of a series circuit and times it by total resistance of that parallel circuit giving me voltage. Then took that voltage and divided it by each individual resistance to find current of each resistor. Anyhow thats all and I would like to ask to scribe again monday.
Final words: I R SRY :)

Circuit Analysis

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THURSDAY JUNE 5, 2008 !

So today in class we did a few things .. THE END just kidding :)



The first thing we did was the last couple of pages for CHAPTER 35 ELECTTRIC CIRCUITS. I'm too shy to tell help you fill in the blanks, so let's move on. The next thing we did was go over SIMPLE CIRCUITS on the white boards. There's alot ot things going on, so if you need the answers ask anyone that was present =).



Here's a little bit of cheat sheet if you are ever doubtful of what belongs to which and who belongs to what and so on so forth.

We also got 2 sheets that deal with COMPOUND CIRCUITS. There are 5 steps to allow us to answer these problems, and they are stated on the first sheet.

PS - I'm not actually shy, I just think you can ask your fellow classmates for the fill-in-the-blanks =)

Wednesday, June 4

ANSWERS!!
Grade 12 Electric Circuits: Resistivity Problems

1) A = ρL/R
= (1.7 x 10^-8)(20) / (0.10)
= 3.4 x 10^-6 m²

A = πr²
r = √(A/π)
= √(3.4 x 10^-6 / π) x 2
d = 2.1 x 10^-3 m

2) L = RA/ρ
= (1.5)(2.5 x 10^-7) / (100 x 10^-8)
= 0.38 m

3) ρ = RA/L
= (0.04)(2.0 x 10^-6 / (5.0)
= 1.6 x 10^-8 Ωm

4) L = RA/ρ
= (3.0)(1.5 x 10^-6) / (100 x 10^-8)
= 4.5 m

5) A = ρL/R
= (5.5 x 10^-8)(0.2) / (0.10)
= 1.1 x 10^-7 m²

A = πr²
r = √(A/π)
= √(1.1 x 10^-7 m² / π) x 2
d = 3.7 x 10^-4 m

Today we corrected Chapter 23: Study Guide Simple Circuits. We were handed Chapter 34: Electric Current (the “fun” thing) and corrected the first page. Lastly, we were given worksheets called Series and Parallel Circuits, and Simple Circuits.

Electric/Magnetic Fields Review

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Tuesday, May 27

Today we corrected the Moving Charges Worksheet. I'm feeling so generous so here's the answers:

4) a) Fe and Fg
b) Positive charge
c) Fe = Fg
mg = Eq
q = mg/E = (3.2 x 10^-15)(9.8) / (19600)
= 1.6 x 10^-18 C

5) B = F/IL
= (0.40) / (20.0)(0.15)
= 0.13 T

6) B = F/IL
= (6.0 x 10^-5) / (1.5)(0.50)
= 8.0 x 10^-5 T

7) F = BIL
= (5.5 x 10^-5)(20.0)(10.0)
= 1.1 x 10^-2 N

8) B = F/qv
= (5.6 x 10^-13) / (1.6 x 10^-19)(2.1 x 10⁵)
= 16.7 T [left]

9) F = Bqv
= (1.2)(1.6 x 10^-19)(8.6 x 10⁴)
= 1.65 x 10^-14 N [right]

10) B = F/qv
= (4.0 x 10^-6) / (1.6 x 10^-19)(5.0 x 10⁵)
= 5.0 x 10⁷ T [into the page]

11) q = F/Bv
= (4.8 x 10^-14) / (5.0 x 10^-2)(2.0 x 10⁶)
= 4.8 x 10^-19 C

12)
a) F = Bqv
= (0.080)(1.6 x 10^-19)(4.0 x 10⁶)
= 5.12 x 10^-14 N

b) R = mv²/F
= (9.11 x 10^-31)(4.0 x 10⁶)² / (5.12 x 10^-14)
= 2.84 x 10^-4 m
13) v = BqR/m
= (5.0 x 10^-4)(1.6 x 10^-19)(0.50) / (9.11 x 10^-31)
= 4.4 x 10⁷ m/s


We also picked up a Review sheet and a worksheet named "The Path of a Charged Particle in a Magnetic Field." Test on Thursday !

Today we learned that in order for an object to move from one distance to another distance above the earth's surface, it needs a certain amount of force which increases its gravitational potential energy (work) to raise a mass from d1 to d2.
Electric potential energy is the same thing as when you place a positive charge between parallel plates. The charge is attracted to the negative plate but in order for the plate to move further away a force must be applied upward. Making a second height other than the first. The area of the shaded region between the two heights reps the increase in electric potential energy.
Calculating Electric potential energy.
PE(1)/(force*height1)=(Force)(distance of charge from negative plate)
=?J (amount of electric potential energy the charge has from the -ve plate)

Now you calculate the PE of charge at a second height
PE(2)=(Force)(second distance of charge from negative plate)
=?J(amount of electric potential energy the charge has from the -ve plate at a second distance)

Now to find the total PE b/w the two positions YOU SUBTRACT PE(1) from PE(2). Now the answer doesnt always have to be positive. Only if point B is further away from the negative plate. If point B is closer than it has a loss of PE.

This can be found from the shaded region of the two distances by multiplying the force of both(only if the same force is applied) by the two distances.


So PE is the work that is done in order to move a charge from one place to another.

ELECTRIC POTENTIAL
Electric potential and PE are very different. Electric potential is the potential energy that a charge has at a given point. Which can be found by this equation ( energy,J)/(charge,C)
The quantity is called a volt.

Potential Difference
To calculate the potential difference of a charge once it has moved to another point you must calculate the Electric potential of the first point V(1)=( energy,J)/(charge,C) and the same for the second point V(2)=( energy,J)/(charge,C). Doing the same thing as in Electric potential energy the Voltage of the first point from the voltage of the second point.

DeltaV=V(2)-V(1)


Now the amount of voltage that has changed in the positions of the charge is EQUAL to the amount of work needed to move the charge from one position to the next.

Now we were given a fun worksheet in which we went over in class and a second problem solving sheet which to be finished by tomorrow.