ECEN 301Discussion #3 – Electric Power1 True Power – The Book of Mormon Alma 37:14, 16, 18 14 And now remember, my son, that God has entrusted you with.

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ECEN 301Discussion #3 – Electric Power1 True Power – The Book of Mormon Alma 37:14, 16, And now remember, my son, that God has entrusted you with these things, which are sacred, which he has kept sacred, and also which he will keep and preserve for a wise purpose in him, that he may show forth his power unto future generations. 16 But if ye keep the commandments of God, and do with these things which are sacred according to that which the Lord doth command you, (for you must appeal unto the Lord for all things whatsoever ye must do with them) behold, no power of earth or hell can take them from you, for God is powerful to the fulfilling of all his words. 18 For he promised unto them that he would preserve these things for a wise purpose in him, that he might show forth his power unto future generations. Mormon 8:15, 16, 22, 26, For none can have power to bring it to light save it be given him of God; for God wills that it shall be done with an eye single to his glory, or the welfare of the ancient and long dispersed covenant people of the Lord. 16 And blessed be he that shall bring this thing to light; for it shall be brought out of darkness unto light, according to the word of God; yea, it shall be brought out of the earth, and it shall shine forth out of darkness, and come unto the knowledge of the people; and it shall be done by the power of God. 22 For the eternal purposes of the Lord shall roll on, until all his promises shall be fulfilled. 26 And no one need say they shall not come, for they surely shall, for the Lord hath spoken it; for out of the earth shall they come, by the hand of the Lord, and none can stay it; and it shall come in a day when it shall be said that miracles are done away; and it shall come even as if one should speak from the dead. 28 Yea, it shall come in a day when the power of God shall be denied…

ECEN 301Discussion #3 – Electric Power2 Lecture 3 – Electric Power

ECEN 301Discussion #3 – Electric Power3 Energy vs. Power Consider filling a pool of water The rate (power) at which the water comes out determines how long it will take to fill the pool The water that has filled the pool represents energy Both pools have the same amount of water (energy) – the rate at which they were filled (power) was different 20 minutes 10 minutes

ECEN 301Discussion #3 – Electric Power4 Energy vs. Power Power: the rate at which energy (electricity) is consumed How fast energy can be generated/delivered to/from an electrical device Energy: the ability to do work EX: a battery, a capacitor, a fuel cell joule (J): electric energy unit. 1 J = the work required to move an electrical charge of 1C through 1V 1 J = the energy required to lift a small apple (102 g) one meter against Earth's gravity. 1 J = the energy required to heat one gram of dry, cool air by 1 degree Celsius. 1 J = eV (electron-volts) 1 J = x10 -7 kilowatt-hour kilowatt-hour (kWh): electric energy unit.

ECEN 301Discussion #3 – Electric Power5 Electric Power Electric Power: the power generated by an active element, or dissipated by a passive element is equal to the product of: voltage across the element and current flowing through it Watt (W): electric power unit. 1 Watt = 1 joule/second (J/s) 1 Watt = 1 volt-amp (VA)

ECEN 301Discussion #3 – Electric Power6 Electric Power Passive sign convention: power dissipated by a load is a positive quantity – the power generated by a source is positive + v ab _ a b i v ba _ a b i + Active Element (source) power dissipated = -vi power generated = vi Passive Element (load) power dissipated = vi power generated = -vi Positive voltageNegative voltage

ECEN 301Discussion #3 – Electric Power7 Passive Sign Convention 1.Choose an arbitrary direction for current flow 2.Label polarities of all active elements (sources) based on direction of current flow 3.Label polarities of all passive elements (loads) based on direction of current flow 4.Compute the power dissipated by each element based on: a)If current flows into the positive terminal of the element, the power dissipated is positive (power is absorbed) b)If current flows into the negative terminal of the element the power dissipated is negative (power is delivered)

ECEN 301Discussion #3 – Electric Power8 Electric Power Example1: find the power dissipated by each element. Ùv a = 8V, v b = 4V, i = 0.1A Load A Load B VsVs

ECEN 301Discussion #3 – Electric Power9 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a clockwise direction of current flow Load A Load B VsVs i

ECEN 301Discussion #3 – Electric Power10 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a clockwise direction of current flow 2.Label polarity of sources according to current flow Load A Load B + _ VsVs i

ECEN 301Discussion #3 – Electric Power11 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a clockwise direction of current flow 2.Label polarity of sources according to current flow 3.Label polarity of loads according to current flow + v a – +vb–+vb– Load A Load B + _ VsVs i

ECEN 301Discussion #3 – Electric Power12 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a clockwise direction of current flow 2.Label polarity of sources according to current flow 3.Label polarity of loads according to current flow 4.Calculate power dissipated in each element + v a – +vb–+vb– Load A Load B + _ VsVs i

ECEN 301Discussion #3 – Electric Power13 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a clockwise direction of current flow 2.Label polarity of sources according to current flow 3.Label polarity of loads according to current flow 4.Calculate power dissipated in each element + v a – +vb–+vb– Load A Load B + _ VsVs i

ECEN 301Discussion #3 – Electric Power14 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a clockwise direction of current flow 2.Label polarity of sources according to current flow 3.Label polarity of loads according to current flow 4.Calculate power dissipated in each element + v a – +vb–+vb– Load A Load B + _ VsVs i

ECEN 301Discussion #3 – Electric Power15 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A 1.Assume a counter-clockwise direction of current flow 2.Label polarity of sources according to current flow 3.Label polarity of loads according to current flow 4.Calculate power dissipated in each element – v a + –vb+–vb+ Load A Load B – + VsVs i

ECEN 301Discussion #3 – Electric Power16 Electric Power Example1: find the power dissipated by each element. v a = 8V, v b = 4V, i = 0.1A – v a + –vb+–vb+ Load A Load B – + VsVs i NB: Power is conserved (power supplied always equals power consumed): P s + P a + P b = 0

ECEN 301Discussion #3 – Electric Power17 Electric Power Example2: find the power dissipated by each element. B E + 10V – – 3V + A D C – 5V + i e = 2A i d =3A

ECEN 301Discussion #3 – Electric Power18 Electric Power Example2: find the power dissipated by each element. B E + 10V – – 3V + A D C – 5V + i e = 2A i d =3A a iaia

ECEN 301Discussion #3 – Electric Power19 Electric Power Example2: find the power dissipated by each element. B E + 10V – – 3V + A D C – 5V + i e = 2A i d =3A i a = 5A Loop1 +va–+va–

ECEN 301Discussion #3 – Electric Power20 Electric Power Example2: find the power dissipated by each element. B E + 10V – – 3V + A D C – 5V + i e = 2A i d =3A i a = 5A Loop2 + 12V – +vd–+vd– NB: since load D and load E share the same node their voltages will be the same

ECEN 301Discussion #3 – Electric Power21 Electric Power Example2: find the power dissipated by each element. B E + 10V – – 3V + A D C – 5V + i e = 2A i d =3A i a = 5A + 12V – + 10V –

ECEN 301Discussion #3 – Electric Power22 Electric Power Example3: The battery supplies a total of 10mW. What is i 3 ? i 1 = 1mA, i 2 = 1.5mA i i i1i1 i2i2 i3i3 4V + _

ECEN 301Discussion #3 – Electric Power23 Electric Power Example3: The battery supplies a total of 10mW. What is i 3 ? i 1 = 0.5mA, i 2 = 1.5mA i i i1i1 i2i2 i3i3 4V + _ +_+_ +_+_ +_+_ Note: we get the same answer but we could have solved for the current like so:

ECEN 301Discussion #3 – Electric Power24 i–v Element Characteristics The relationship between current (i) and voltage (v) at the terminals of a circuit element defines the behaviour of that element within the circuit + _ i v If a voltage is imposed across the terminals of the element the voltage itself (v) and the current that will flow (i) form a unique pair of values. measuring the current while the voltage varies is known as the i–v characteristic with the i–v characteristic voltage (current) can be determined from a given current (voltage) power can also be determined

ECEN 301Discussion #3 – Electric Power25 i–v Element Characteristics i–v characteristic can be given as: a function a graph

ECEN 301Discussion #3 – Electric Power26 i–v Element Characteristics Power generated vs. power dissipated quadrants Generated (sources) Generated (sources) Dissipated (loads) Dissipated (loads)

ECEN 301Discussion #3 – Electric Power27 i–v Element Characteristics Elements that source and dissipate power Photovoltaic – solar cells, photodiodes, photodetectors Dissipate Generate

ECEN 301Discussion #3 – Electric Power28 i–v Element Characteristics Ideal DC current source 3A source Ideal DC voltage source 6V source + _ VsVs IsIs