Tim Cinel

Q1

a) List four reasons why engineering students need to learn the fundamentals of electrical engineering.

1. Design electronics
2. Understand how circuits work
3. Know how to measure and quantify performance
4. More advanced theories require fundamental knowledge

b) List eight subdivisions of electrical engineering

1. Control
2. Communication
3. Power
4. Signal Processing
5. Optics
6. Computation
7. Electromagnetic
8. Electronics

Q2

Carefully defined or explain the following terms in your own words:

a. electrical current

The motion of charged particles through a medium.

b. voltage

The amount of energy taken to move charge between two points. Quantitatively, voltage is a measures of joules per coulmb.

c. open switch

Part of a circuit that disrupts continuity, breaking the circuit.

d. closed switch

Part of a circuit which can break the circuit but is connected, maintaining continuity.

d. direct current

Charge flowing through a medium at a fairly consistent rate.

e. alternating current

Charge moving back and forth within a medium generally in a sinusoidal manner.

Q3

The charge of an electron is -1.60 x 10-19 C. A current of 1A flows in a wire carried by electrons. How many electrons pass through a cross section of the wire each second?

Ans = n(e) * I

Electrons per coulomb of charge:

1/(1.6×10^-19) * 1 = 6.25×10¹⁸

Q4

The circuit element shown in figure below has v = 12V and i_ba = -2A. What is the value of v_ba? Be sure to give the correct algebraic sign. What is the value of i? is energy delivered to the element or taken from it?
Vba = -12V

I = -2A

taken

Q5

The net charge through a cross section of a circuit element is given by q(t) = 2 + 3t C.

Find the current through the element.

I = sqrt(2² + 3²)

= sqrt(12) Amp

Q6

A certain lead acid storage battery has a mass of 30kg. Starting from a fully charge state,

it can supply 5 amperes for 24 hours with a terminal voltage of 12V before it is totally

discharge.

a) If the energy stored in the fully charged battery is used to lift the battery with 100-

percent efficiency, what height is attained? Assume that the acceleration due to

gravity is 9.8m/s and is constant with height

E = 12*5*24*60*60 Joules

= 5.4×10⁶ Joules

E = mgh

h = E / mg

= 5.4e6 / ( 30 * 9.8 )

= 20 km

b) If the energy stored is used to accelerate the battery with 100-percent efficiency,

what velocity is attained?

E = (1/2)mv²

v = sqrt(2E/m)

= sqrt(360000)

= 600 ms^-1
c) Gasoline contains about 4.5 x 10⁷ J/kg. Compare this to the energy content per

unit mass for the fully charged battery.

E/m_batt = 5.4e6 / 30

= 1.8e5 J/kg

E/m_petrol = 4.5e7 J/kg

Q7

Suppose that the cost of electrical energy is $0.12 per kilowatt hour and that your

electrical bill for 30 days is $60. Assume that the power delivered is constant over the

entire 30 days. What is the power in watts? If this power is supplied by a voltage of 120V,

what current flows?

E = (60 / 0.12) kWh

= 500 kWh

= 1.8e9 J
P = E / t

= 1.8e9 / 60*60*24*30

= 694.44 W

P = VI

I = P/V

= 694.44 / 120

= 5.8 A
Part of your electrical load is a 60W light that is on continuously. By what percentage can

your energy consumption be reduced by turning this light off?

% = 100*(Plight / Ptotal)

= 100*(60 / 694.44)

= 8.64%

Tags: Circuit Theory, ohm's law, Physics, solution, tutorial

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