Paul confronted him and he said it’s ok because he’s a “photographer.”

We think he was upto something suss, so here’s a photo of him:

Honestly,

I’m Not A Thief, I’m A Photographer

Generated by my Lovely Caption Maker

(Sorry about quality, taken with my phone)

Q6

Considering the circuit in Fig.Q6, calculate the voltage across the 50 Ω resistor and the power supplied by the sources.

No matter how I try it, I can’t get the right values out of this mesh analysis… What am I doing wrong?

Mesh #1 Mesh #2

Solve V1A and i2

(Ans: 75 V, P1A = 175 W, P100V = 50 W)

Circuits with controlled sources

Q7

Calculate the current through the controlled voltage source in Fig.Q7.

I’ve tried this with mesh and got the wrong answer, so I tried node voltage analysis which still doesn’t work… What am I doing incorrectly?

Node #1 KVL (Leaving)

Node #2 KVL (Leaving)

Node #3KVL (Leaving) Substitute V3 into Equations

Solve ix Solve

[Ans: 3 A downward]

Q8

Calculate the equivalent resistance as seen from the terminals a and b for the network shown in

Why does a 1A connection between a and b allow me to solve the equivalent resistance? Where is this in our lecture notes?

{Hint: Connect a 1A current source between terminals a and b. Then solve for the mesh currents using the mesh current analysis. Having calculated the mesh currents, find the voltage across a-b.

This voltage is equal to the value of the equivalent resistance between terminals a and b.}

[Ans: -15Ω]

Ideal Independent Voltage and Current Sources

Q1

In the circuits shown in Fig.Q1(a) & Q1(b), calculate the power absorbed by the three circuit elements. ( The voltage source, current source and the resistor.) Confirm that your answers satisfy the principle of conservation of energy.

Fig Q1 a

Fig Q1 b

(Ans: (a) PI = -60 W, PV = +15 W, PR = +45 W.(b) PI = -10 W, PV = -6 W, PR = +16 W.)

Q2

(a) determine the ammeter and voltmeter readings.

A = 2 A

(b) What are the power supplied by the sources and the power absorbed by the resistor?

Sources

Resistors

(Ans: (a) -2 A, 20 V, (b) Current source supplies 40 W; Voltage source absorbs 24 W; and the resistor absorbs 16 W.)

Q3

In the circuits shown in Fig.Q3, determine the current flowing through the voltage source and the voltage across the current source. Mark your reference direction of the current and the voltage polarities. Hence calculate the power supplied by both sources and the power absorbed by the two resistors.

(Ans: IV = 2 A down, 34 V, top+, Power supplied by V source = -36 W, Power supplied by I source = +136 W, Power absorbed by 4 Ω and 9 Ω are 64 W and 36 W respectively.)

Q4

In the circuit shown in Fig.Q4, calculate the source current iS. Check the power balance in the circuit.

[Ans: iS = 0.5 A; Power absorbed, P16V =-8 W, P4Ω = 1 W, Pcont= 7 W]

Q5

In the circuit shown in Fig.Q5, calculate the currents supplied by the independent and dependant voltage sources.

[Ans: iS = 2 A, iY = -1 A]

Q6

In the circuit shown in Fig.Q6,

Calculate the independent

source voltage.

[Ans: VS = 18 V]

Q7

In the circuit shown in Fig.Q7, calculate the current IS of the independent current source.

[Ans: IS = 1 A]

Q8

In the circuit shown in Fig.Q8, determine the voltmeter reading. Pay attention to the + polarity of the voltmeter.

(Ans: +4 V)

Q9

In the circuit shown in Fig.Q9, determine the voltage across the current source and the ammeter reading.

(Ans: 9 V, 3 A)

Q10

In the circuit shown in Fig.Q10, calculate (a) the voltage across the 24 Ω resistor, (b) the current flowing into the 9 Ω resistor and (c) the current flowing into the 4 Ω resistor.

(Ans: (a) 24 V, (b) 2 A, (c) 1.5 A )

Q11

In the circuit shown in Fig.Q11, calculate the current supplied by the 40 V source.

(Ans: 2 A)

Reference directions of currents and numerical values, Kirchhoff’s Current Law

Q1

With the reference directions of currents shown in Fig.Q1, calculate the

numerical values of the missing current I1: (Ans: 2 A, 6 A, -2 A)

a) Outgoing = 0 -6 -4 + 8 + I = 0 2 = I

b) Outgoing = 0 15 – 7 -2 -I = 0 I = 6

c) Outgoing = 0 -6 - 4 + 12 + I = 0 I = -2

Q2

Mark the reference direction and calculate the numerical values of the

missing currents in the following parts of circuits in Fig.Q2:

(Ans (a) 8 A, 5 A, 7 A, 4 A, (b) 6 mA, 3 mA, 5 mA )

Kirchhoff’s Voltage Law

Q3

With the voltage polarity marked in the circuits below, calculate the

numerical values of the unknown voltages:

(Ans: (a) 42 V, -28 V, (b) 14 V (c) 8 V, 6 V, 4 V)

V₁ = 42V

V₂ = -28

V = 42 – 28

V = 14V

-12 + 6 – 2 + V1 = 0 V1 = 8V

-12 + 6 + V2 = 0 V2 = 6V

-12 + 8 + V3 = 0 V3 = 4V

Resistor

Q1

A resistor is being tested using the circuit shown in Fig.Q1 in which a variable dc power

supply delivers current to the resistor.

(a) With one setting of the power supply the ammeter reads 2 mA while the voltmeter reads 4.4 V. Calculate the resistance.

R = 2.2 kΩ

(b) The current is then increased to 5 mA. What is the expected voltmeter reading?

(Ans: (a) 2.2 kΩ, (b) 11 V)

V = 11V

Q2

In the circuit shown in Fig.Q1, (a) calculate the power dissipated in the same 2.2 kΩ

resistor when the ammeter reads 2 mA.

W = 8.8mW

(b) If the resistor power rating is known to be 0.25 W, what is the maximum voltage to which you can raise the supply voltage safely?

(Ans: (a) 8.8 mW, (b) 23.4 V)

Let P = 2.5*10² W

Q3

A 22 kΩ and a 2.2 kΩ resistor are connected in series and are used as voltage divider in the circuit shown in Fig.Q3. Both resistors have the same power rating of 0.25 W.

(a) When the supply voltage is 50 V, what are the ammeter and voltmeter readings?

(b) Can you raise the supply voltage to 100 V safely? If not, why?

22 Ohm Power > 0.25 W

What is the highest voltage to which the supply voltage can be raised?

(Ans: (a) 2.1 mA, 4.5 V, (b) No, 22 kΩ overheats (c) 81.5 V)

Q4

A 24 kΩ, 0.25 W and a 56 kΩ, 0.25 W resistor are connected in parallel as in the circuit shown in Fig.Q4.

(a) The power supply voltage is adjusted to 42 V. What are the expected readings of V, A1, A2 & A3?

(b) What is the highest voltage to which the supply voltage can be raised?

V_max = 77.5 V

(Ans: (a) 42 V, 1.75 mA, 0.75 mA & 2.5 mA (b) 77.5 V)

]]> http://www.timcinel.com/2008/eeet-2249-tutorial-3/feed http://www.timcinel.com/2008/eeet-2249-tutorial-2 http://www.timcinel.com/2008/eeet-2249-tutorial-2#comments Sat, 13 Sep 2008 06:03:43 +0000 Tim Cinel http://www.timcinel.com/?p=62

Charge, Current, Voltage, Power and Energy Charge & Current

Q1

A wire carries a constant current of 10 mA. How many coulombs pass a cross

section of the wire in 30 s?

n(C) = i*T

= 10e-3 * 30

= 3e-1 C

(Ans: 0.3 C)

Q2

The charge entering the terminal of a device is given by q(t) = k₁t + k₂t² C.

It is observed that i(0) = 5 A and i(4) = -3 A. Find the constants k₁ and k₂ .

(Ans: k₁ = 5 A and k₂ = -1 A/s ).

Q3

The time varying current in an element is given by:

Calculate the total charge that has entered the element in (a) t = 6 s and (b) t = 10 s.

(Ans: q(6) = 8 C and q(10) = 5 C)

a)

q = (5-2)(3) + (6-5)(-1)

q = 8 C

b)

q = (5-2)(3) + (9-5)(-1)

= 5 C

Q4

In an electroplating bath silver ions transport charges from the positive silver

electrode to the negative electrode, which is an object being silver plated. When the

positive silver ions arrive at the negative electrode the charges on the silver ions are

neutralized and the silver is deposited on the surface of the negative electrode.

Given that 1.118 mg of silver transports 1 C of charge, determine the weight of silver

deposited when 5 A current is passed through the bath for 30 minute.

q = i*t

= 5 * 30 * 60

= 9000 C

m = 9000 * 1.118e-3

= 10.06 g

(Ans: 10.06 grams)

Voltage, Energy & Power

Q5

A and B are two points in a circuit with one source of energy. It is determined that to

move a +2 coulomb of charge from A to B, the source spends 10 joules of energy in 4

seconds.

(a) What is the voltage across the points A and B? (Mark polarities and give the

numerical value) (Ans: Polarities: +A and –B, voltage 5 V)

A+ B- 5V

(b) Assuming steady flow of charges over a 4 second period, calculate the power spent

and the current in the circuit? (Ans: power = 2.5 W, current = 0.5 A)

P = W/t

= 10/4

= 2.5 W

I = q/t

= 2/4

= 0.5 A

Q6

Four AA alkaline batteries are used in the 6 V power pack of a radio set. Each

battery has a useful energy of 50 watts seconds (joules). If the radio takes a steady current

of 2 mA from the power pack, how long can the batteries last? (Ans: 4.63 hours)

w = qV

q(total) = W/V

= 50 / 6

= 8.33 C (per battery)

= 33.32 C (total)

I = q/t

t(total) = q / I

= 33.32 / 2e-3 sec

= 4.63 hours

Q7

The terminals of a certain battery are labeled a and b. The battery voltage is V_ab =

12V. To increase the chemical energy stored in the battery by 600J, how much charge

must be move through the battery? Should the electrons move from a to b or from b to a?

(Ans: 50C, b to a)

W = qV

q = W/V

= 50 C

Battery with Vab = 12V implies b is negative and a is positive. To charge, charge must go from b to a, so electrons must go from b to a.

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%

Key Points:

• 80% of electricity bills
  is not for electricity
• 75% of energy is wasted
  before leaving power plants
• Centralised power plants
  waste large amounts of water
• Energy can by supplied
  25% of current cost
• Decentralised energy is
  more economically efficient

Do you ever consider what components of your power bill add up to the rate that you’re paying? According to innovative British engineer Allan Jones, about 20% of the rate accounts for actual electricity. The majority of the rate is to pay for distribution, power loss and government grid fees. Jones recognised these financial and fuel inefficiencies and proposed a solution, local decentralised power, that managed to save a council 80% in energy costs and is now being implemented in London.

Jones planted the seed of decentralised power in 1990, when was a senior officer in council for the borough of Woking and submitted a report on global warming - long before it became an epidemic. The council moved very quickly and set up a revolving fund for the environment. The phenomenon started from there.

The primary cause of climate change is centralised energy. Centralised energy is energy generated by big power stations and delivered over long distances to many locations. Most centralised power plants use Coal as the fuel, which is about five times more polluting than gas. Most centralised power plants also waste two thirds of their energy as unused heat, in fact they waste large amounts of water to dissipate this heat. Considering a significant portion of consumer energy is spent on heating, the wasted heat seems rather unfortunate.

With such serious inefficiencies Allan Jones and the council of Woking recognised the opportunity to revolutionalise energy supply, the revolution of decentralised energy. Decentralised energy is a much more efficient of distribution electrical and heat energy through municipalities.

Decentralised energy is generated locally, that is, a small power plant is located close to or within the town or city it supplies. Both the electricity and heat are distributed – this is called ‘combined energy’. This way most of the energy is used, rather than most of it being wasted through heat dissipation. Much less electrical energy is lost through transmission because the distances are so short. Additionally, the heat that is usually wasted can be used to power heating inside buildings, hot water systems, and even refrigeration and cooling through a technology called ‘heat fired cooling’ - operations that usually require energy.

New infrastructure was required in order to generate and then deliver the electricity and heat, requiring a hefty amount of capital. However, the economic advantage of decentralised power made this easily feasible for Jones. £250 thousand was provided by the council Woking, a town with a population of 100,000. Jones had one condition, “Providing you allow me to recycle the financial savings from the energy bills, I won’t need any more money from you…”. Jones didn’t need any more money for the implementation.

The council of Woking saved about 1.2 billion a year on energy costs after the implementation of decentralised power while profiting from the joint venture energy services company established by Jones. Economically and environmentally Woking has done very well out of decentralised energy.

Jones believes that big energy companies can either interpret the current energy climate as a threat or as an opportunity. They can be large dinosaurs, refuse to change and become extinct or they can adapt by tackling climate change head-on. Centralised energy must be changed, Jones believes, since it is responsible for around 75% of emissions for some large cities.

Source: Allan Jones: getting off the grid - ABC Radio National’s “Saturday Extra”, 26 July 2008

The proposed framework for Australia’s solution to carbon emissions penalises the production of carbon emissions. Since most other countries are not implementing such schemes, this makes locally produced products more expensive. In reaction to higher domestic products, consumors will buy more imported products. Not only does this result in the collapse of Australian industries but also will not decrease carbon emissions – only shift them overseas where carbon emissions are not capped.

Internally carbon pricing consumed goods and services is a possible solution to this problem with many advantages. Charging the Australian consumer for the carbon emissions of the things they use can curb global carbon emissions while keeping economically competitive globaly. Carbon emissions are curbed because whether a product generating carbon emissions came from China, Fiji or Australia, more tax is charged for more carbon emissions – slowing demand and also funding environmental rehabilitation. On the other hand, Australia remains competitive because export prices remain on a level playing field, untaxed when sold offshore.

Keep in mind that the production model would allow Australia to import aluminium, steel and other energy-intense products from carbon-inefficient sources like China, no carbon prices would be passed onto the consumer. Although these producing countries are releasing the emissions, they are certainly not the only consumers. Wealthy western countries account for major exports from the highest emitting countries so shouldn’t they be penalised for the emissions their demand has produced?

This approach to solving global warming though carbon capping shares common disadvantages with the currently-accepted carbon production tax approach: A hit to Australians’ lifestyles. We might have to use more public transport or cycle more while being more conscious of lights and electrical appliances we’ve left on. Even if global warming isn’t real, sustainability is a great thing.

… from a radio interview with:

Geoff Carmody
Principal Geoff Carmody Associates Co-Founder Access Economics

Martijn Wilder
Baker & McKenzie

Entire Source: ABC Radio National Saturday Extra 2008-08-09 (30:00 - 45:00)
Snippet Mirror:  TimCinel.com [Saturday Extra 2008 08 09 - Consumer-Based Carbon Tax]

This made me laugh. According to this report on New Zealand Immigrants from the ABC, the rate of residents leaving New Zealand is growing at an alarming rate. The funny part is that most of the growth is due to residents that weren’t born in New Zealand!

New Zealand residents are able to come to Australia to work and live without even applying for a visa. If the trend continues, Revenue Minister Peter Dunne says New Zealand will become a “transit lounge” for immigrants.

Even immigrants know that Australia is better than New Zealand.

Have a listen to the funniest but creepiest prank call made on Fox FM’s Matt & Jo show - the segment is about five minutes into it:

Hosted on Fox FM: Creepy Phone Sex Line

All I can say is … “Cock a doodle -doooo - hooo - hooo”

Grettle Would Smile At That