**The Exam**

FE Chemical

FE Civil

FE Electrical and Computer

FE Environmental

FE Industrial

FE Mechanical

FE Other Disciplines

**Work,**

Energy,

Power,

Heat

Energy,

Power,

Heat

**Inductance**

Measured in henries (H)

H = (m^2*kg)/C^2

Opposes change in current

Looks like short circuit as frequency >> 0 Hz

looks like open circuit as frequency >> ∞ Hz

Work on a Charge in Electric Field

**General FE Exam Info**

**Calculator Policy...**

Statistics..........

Other Info..........

Statistics..........

Other Info..........

Charge, Energy, Current, Voltage, Power

**Electrical and computer Fe Exam**

-Engineering Sciences -

-Engineering Sciences -

**By:**

Aaron Craven

Irina Tyx

Aaron Craven

Irina Tyx

Electrical and Computer

1. Mathematics

2. Probability and Statistics

3. Ethics and Professional Practice

4. Engineering Economics

5. Properties of Electrical Materials

6. Engineering Sciences

7. Circuit Analysis (DC and AC Steady State)

8. Linear Systems

9. Signal Processing

10. Electronics

11. Power

12. Electromagnetics

13. Control Systems

14. Communications

15. Computer Networks

16. Digital Systems

17. Computer Systems

18. Software Development

Engineering Sciences

6-9 Questions

A. Work, energy, power, heat

B. Charge, energy, current,

voltage, power

C. Forces (e.g., between charges,

on conductors)

D. Work done in moving a charge in an

electric field (relationship between

voltage and work)

E. Capacitance

F. Inductance

Forces

The work done by an external agent in moving a charge Q in an electric field from point p1 to point p2 is:

Capacitance: Unit [F]

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Hewlett Packard

Texas Instruments

HP 33s

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TI-30Xa

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TI-30Xa SE

TI-30XS Multiview

TI-30X IIB

TI-30X IIS

TI-36X II

TI-36X SOLAR

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Angle Conversions

Vectors and Matrices

Simultaneous Equation Solver

Time:

6hr appointment

5hr 20min for exam

Questions:

110 Multiple Choice

2min 54sec per Question

WDTM:

Pass Rate:

82% of First-Time Takers for Electrical and Computer Exam

Results:

Results in 7-10 Days

Pass or Fail

Ambiguous Passing Score

Diagnostic Report if you Fail

During The Exam

So What's Next?

Charge

Energy

Current

Voltage

Power

Sample Problem

Fundamental property of subatomic particles

Electron (-1.6x10^-19 Coulombs)

Proton (+1.6x10^-19 Coulombs)

Conservation of charge

Measured in Joules

Not to be confused with work

Force F exerted E joules of energy on mass M which resulted in W joules of work done on the mass

Current is the movement of charge

Opposite direction of electron flow

Defined as charge per time

SI units coulombs/second (ampere)

i(t) = dq(t)/dt

I = dQ/dt

J

= p

v

Measured in watts

Energy per time

W = J/s = N*m/s = kg*m^2/s^3

P = IV

Measured in volts

Strength of electric field

Difference in electric potential between two points

Work required to move 1 coulomb from A to B

V = IR = P/I

Q. A current of 5A flows through a wire with a diameter of 3mm. What is the average number of electrons per second that passes through any given cross sectional point along the wire?

A. 8.0 x 10^19

B. 3.1 x 10^-19

C. 1.6 x 10^19

D. 3.1 x 10^19

1 A = 1 C/s (def)

5 A = 5 C/s

1 e = 1.6x10^-19 C (def)

5 C/s x 1/(1.6x10^-19) e/C

=

3.1x10^19

electrons/second

Answer is D.

Wire diameter is irrelevant.

Sample Problem

What is the work required to move a charge of +5 C a distance of 8 m in a uniform electric field of 100 V/m?

A. -800 J

B. 0 J

C. -4000 J

D. 4000 J

Inductance Equations

Sample Problem

What is the equivalent inductance of the circuit below?

A. 5 H

B. 20 H

C. 23.75 H

D. 25 H

L4 and L3 are in series with L2 and L1 which are in parallel.

Leq = L4 + L3 + L2//L1

= 15 + 5 + (10)(10)/(10 + 10)

= 20 + 100/20

= 20 + 5

=

25 H

Answer is D.

5 hours, 20 minutes to complete

raise hand for more

reusable booklets

scheduled break.......

unscheduled breaks....

after submitting ~55 Q's

25 minutes

may take fewer minutes, but no time incentive

may leave the building

return on time!

clock will automatically resume

may take anytime

amount of time will be deducted from testing time

access to personal items

beverages, food, and items on Comfort Aid List

EIT (Engineer in Training)

Work for at least 4 years

Take the P.E. exam

managerial positions

higher salary

new job opportunities/responsibilities

bid for gov't contracts

own a firm

consult

offer expert witness testimony

stamp and seal designs

Work

defined as "transfer of energy"

physics says, "work is done on object when you transfer energy to that object"

if one object transfers (gives) energy to a 2nd object, then the 1st object does work on the 2nd object

W = Fxd

Units: N*m, or Joules

Energy

defined as capacity for doing work

Kinetic Energy = energy of motion

Potential Energy = stored energy

Forms

Solar radiation

infrared, radio, UV, etc

Atomic/Nuclear Energy

fission: neutron splits atom's nucleus into smaller pieces

fusion: two nuclei joined together under millions of degrees of heat

Electrical Energy

generation/use of electric power over period of time

kWH, GWh

Chemical Energy

potential energy related to breaking/forming of chemical bonds

Mechanical Energy

Heat Energy

Kinetic: Ek = 1/2*m*v^2

Potential: U = 1/2 *k*x^2

Units: N*m or Joules

Power

Work done in a unit of time

measure of how quickly work can be done

Unit

SI: Watt = 1 Joule/1 second

English: 1 HP = 735.7 W

Example:

if we are using 1 kW of power, a kWh of energy will last one hour

Heat

heat is a form of energy, or thermal energy

heat energy is transferred as a result of temperature differences

Heat flow

Watt = 1 J/s of heat flow

Sample Problem

A 3500 kg car traveling at 65km/h skids and hits a wall 3 seconds later. The coefficient of friction between the tires and the road is 0.60.

Assuming that the speed of the car when it hits the wall is 0.20 m/s, what energy must the bumper absorb in order to prevent damage to the car?

(A) 70 J

(B) 140 J

(C) 220 J

(D) 360 kJ

Solution

Use the kinetic energy equation

Ek = 1/2 * m * v^2

= (1/2)(3500 kg)(0.2 m/s)^2

= 70 J

Answer is (A).

Force on test charge Q in electric field E:

F = QE

Force experienced by point charge 2 (Q2) in an electric field E, created by point charge 1 (Q1):

Sample Problem

Solution!

represents the ability to store charge

the greater the capacitance, the greater the charge stored

Capacitance of two parallel plates of equal area A, separated by distance d. Where ϵ is the permittivity of medium separating the plates

Current in a capacitor

can change instantaneously

Voltage

cannot change instantaneously

will change as the capacitor integrates the current to produce a voltage that opposes the change

Total energy (in J) stored in capacitor

Capacitors in Series:

like resistors in parallel

Cs = 1 / (1/C1 + 1/C2 +...+ 1/Cn)

Capacitors in Parallel:

like resistors in series

Cp = C1+C2+...+Cn

RC Transient Circuits

time constant

τ = RC

time it takes voltage to reach 63.3% of its steady state value

in general, transient variables will have reached their steady-state values after 5 time constants (99.3% of steady-state value)

Sample Problem

What are the voltage Vx and current Ix values through the 5Ω resistor in the center leg after a long time?

(A) 2V; 1A

(B) 2V; 3A

(C) 5V; 1A

(D) 5V; 20A

Solution!

after t=5τ, the capacitor acts like an open circuit.

Using Ohm's Law for Ix, the answer is C (5V; 1A)

Congratulations, you passed!