Fall 2001Circuits I1 Chapter 1 - Basic Concepts 1.1System of Units 1.2Basic Quantities 1.3Circuit Elements

Fall 2001Circuits I2 Reading Assignment: Chapter 1, Pages 1-11 Homework End-of-Chapter Problems: P1.13, P1.18, P1.21 (PLUS WebCT Problems)

Fall 2001Circuits I3 Electrotechnology pervades our day-to-day lives:  electrical appliances  Refrigerators, stoves, stereos, hairdryers,...  communication devices  telephones, faxes, computer and data networks  computers  main frames, personal (PCs), workstations, embedded systems, PLCs  instrumentation  meters, measurement systems, controls “Today we live in a predominately electrical world.”

Fall 2001Circuits I4 System of Units The international system of units (SI) is composed of the following basic units of measure:

Fall 2001Circuits I5 Standard SI Prefixes Standard prefixes are employed with SI units to represent powers of ten. These prefixes, sometimes referred to as engineering notation, are used extensively in representing the values of various circuit parameters kilo (k) mega (M) giga (G) tera (T) pico (p) nano (n) micro (  or u) milli (m)

Fall 2001Circuits I6 TI-85 MODE Function To Select Engineering Notation

Fall 2001Circuits I7 Example - Using Engineering Notation If the local power company charges $0.12 per kWh (kilowatt-hour) of energy used, how much would it cost to illuminate a 100-Watt lamp for an entire year? Solution: TI-85 Screen (Wh) (kWh)

Fall 2001Circuits I8 Basic Quantities What is Electricity? - A natural force (like wind or water flow) which can be harnessed for some beneficial purpose. What causes electricity? Just as wind is due to the movement of air molecules, electricity is caused by the movement of charged particles (i.e., electrons). The study of electricity has been going on since about 600 B.C. when it was noticed that amber, when rubbed, acquired the ability to pick up light objects (we call this static electricity).

Fall 2001Circuits I9 Definitions Charge is an intrinsic property of matter (like mass and size). Symbol 1 : q or QUnit: coulomb (C) The elemental unit of charge is the charge of one electron, e: e = x C (1 proton = x C) Current is a measure of the amount of charge flowing past a point in a given amount of time. Symbol 1 : i or IUnit: ampere (A) 1A  1C/s (coulomb per second) 1 Lowercase is used to indicate a variable (time-dependent) quantity while upper case indicates a constant value.

Fall 2001Circuits I10 Current has magnitude and direction. Electrical Conductor Segment 2A Conventional Current Flow ---- Electron Flow (The flow of electrons is analogous to the flow of water molecules in a pipe or hose.) Note: 2A = 2C per second = (2/1.6022E-19) electrons per second 2A  12,500,000,000,000,000,000 electrons per second

Fall 2001Circuits I11 Typical Current Magnitudes Current in Amperes (A)

Fall 2001Circuits I12 Direct Current (dc) Versus Alternating Current (ac) DC signals are constant with respect to time: t IxIx 2A AC signals vary with respect to time: t i(t) 2A -1A periodic ac signal t i(t) = 2e -at 2A nonperiodic ac signal

Fall 2001Circuits I13 Charge - Current Relationships Current is proportional to the rate of change of charge. Charge is increased or decreased depending on the area under the current function.

Fall 2001Circuits I14 Example 1 Given:q(t) = 12t C Find:i(t) Solution: Since current is proportional to the rate of change of charge, we take the derivative of q(t) to find i(t): i(t) = q(t) i(t) = 12 A Since the slope (rate of change) of q(t) is constant at 12 Coulombs per second, the current is constant.

Fall 2001Circuits I15 Example 2 Given: The current entering an element is described by the function in the graph shown below. t ( seconds ) i(t) Find: The amount of change in the charge of the element over the interval 0 to 3 seconds.

Fall 2001Circuits I16 Example 2 - Solution To find the change in the charge of the element, we need to find the area under the function i(t) over the interval 0 to 3 s. This can be done two ways for this example: by integration or by using area formulas for rectangles and triangles. t i(t) Note: 5C is the increase in the charge of the element. We must know the initial charge to determine the total charge of the element.

Fall 2001Circuits I17 Definition - Voltage As stated earlier, current is the flow of charge through a conductor. Voltage is the force required to move the charge through the conductor. Hence, voltage is also referred to as electromotive force and potential difference. + - Battery Lamp i (current flow) + V ab - + V ab - i b a

Fall 2001Circuits I18 Battery Lamp i (current flow) + V ab - + V ab - The current from the battery flows through a voltage rise since the battery is supplying energy. There is an equal voltage drop across the lamp since the lamp is absorbing the energy supplied by the battery. This is similar to the counter force created by friction when pushing an object up an incline. F ab b a

Fall 2001Circuits I19 Electromotive force is measured in volts. If one joule of energy is required to move one coulomb of charge through an element, then there is one volt (V) of potential difference across the element. Like current, the voltage across an element may be positive or negative. The assignment of a sign to a voltage is somewhat arbitrary, but usually represents whether potential difference is above or below some reference potential.

Fall 2001Circuits I20 As we traverse the circuit in a clockwise direction, there is a voltage rise across the battery; we could assign this a “positive” sign. Since we experience a voltage drop across the lamp, this would be assigned the opposite sign or a “negative” value. We could just as correctly assign all voltage drops a positive sign and, therefore, all voltage rises a negative sign. + Battery Lamp i + 6 V V - “0” volts (ground) (6 V above ground)

Fall 2001Circuits I21 Power and Energy Energy is the capacity to do work. For example, energy is stored in a battery and, when connected to a “load” such as a lamp, there is a continuous exchange of energy from the battery to the load. (There is also a transformation from energy stored in a chemical form to energy emitted in the form of heat and light.) Typically, electrical energy is measure in joules (J) or watt-hours (W-h). (1W = 1J/s) Battery Lamp i + V ab - + V ab -

Fall 2001Circuits I22 Battery Lamp power flows from the source to the load +V-+V- +V-+V- i i p = -vi Supplies Power p = +vi Absorbs Power (Passive sign convention) Power is a measure of how much work is done per unit time, or, in other words, a measure of the rate at which energy is transferred (supplied or absorbed). There is always a conservation of energy (power). That is, the battery supplies exactly the amount of power consumed by the load, no more no less. Typically, electrical power is measured in watts (W).

Fall 2001Circuits I23 Mathematical Relationships In electrical terms: In general: Using previous definitions:

Fall 2001Circuits I24 The total energy absorbed or supplied is equal to the area under the power function in the interval t 0 to t. If the area is positive, then energy is absorbed by an element; if the area is negative, the element is supplying power.

Fall 2001Circuits I25 Example 3 - The current in a typical lightening bolt is 20KA, and the bolt lasts typically for 0.1 seconds. If the voltage between the earth and clouds is 500MV, find (a) the total charge transmitted by the bolt and (b) the total energy released by the bolt. Solution:(a) (b) Note, If V is constant, Then, A typical automobile battery is rated at 12V and 100 amp-hours (Ah). 100Ah = (100 C/s)  (3600s) = 3.6  10 5 C  12V  3.6  10 5 C = 4.32  10 6 J (MJ) The lightening strike has as much energy as 230,000 batteries!

Fall 2001Circuits I26 Example 4 - (a) Find the power absorbed by the element shown and (b) the total energy absorbed over a 10-second interval. (b) 10A + 4v - Solution: p = vi =4v x 10A = 40w (a)

Fall 2001Circuits I27 Example 5 - Find the power absorbed or supplied by the elements shown below Solution: (a) p = -vi = -2v x 4A = -8w (element-a is supplying 8w of power) 4A - 2v + (a) + 2v - (b) -2A (b) p = -vi = -2v x (-2A) = +4w (element-b is absorbing 4w of power)

Fall 2001Circuits I28 Example 6 - Determine the unknown voltage or current. Solution: (a) -20w = -V x 5A V = 4v 5A - V=? + (a) - 5v + (b) I=? P = -20wP = 40w (b) 40w = -(5v x I) I = -8A

Fall 2001Circuits I29 Circuit Elements Circuits (electrical systems) consist interconnected devices. Electrical devices can be categorized as either a source or a load. A source can be either independent or dependent, a voltage source or a current source. Independent voltage sources provide power at a constant voltage Independent current sources provide power at a constant current. The voltage level of a voltage source is not affected by the amount of power demanded from it. Similarly, the current level of a current source is not affected by the amount of power demanded from it.

Fall 2001Circuits I30 Independent Voltage Sources: +V-+V- Constant (battery) LOAD I + v(t) Time-varying LOAD i(t) p v V Since p = vi, as the demand for power increases, the current, i, increases because v is constant.

Fall 2001Circuits I31 Independent Current Sources: p i I Since p = vi, as the demand for power increases, the voltage,v, increases because i is constant. + v(t) or V - LOAD i(t) or I

Fall 2001Circuits I32 Dependent Source (Controlled) Sources) An independent source provides power to a circuit at a constant voltage or current, and the level of that voltage or current is unaffected by the circuit to which it is attached. Generally, independent sources are used to represent devices such as batteries or generators. A voltage or current associated with a dependent source is controlled or affected by the circuit to which it is attached. Dependent sources, also called controlled sources, are used to model electronic devices, sensors and other devices which are influenced by some physical or electrical parameter. For example, a device called a thermocouple generates a small voltage (in the millivolt range) which is proportional to temperature. So, a thermocouple can be thought of as a temperature-controlled voltage source. That is, v(t) = KT, where T is the temperature and k is a constant,

Fall 2001Circuits I33 Dependent Sources There are four types of controlled sources: voltage-controlled voltage sources (VCVS) current-controlled voltage sources (CCVS) voltage-controlled current sources (VCCS) current-controlled current sources (CCCS) +vo-+vo v =  v o - VCVS ioio v = ri o - CCVS

Fall 2001Circuits I34 Dependent Sources VCCS +vo-+vo- i = gv o CCCS i =  v o ioio

Fall 2001Circuits I35 Example 7 - Given the two networks shown below we want to determine the outputs. Solution: (a) V o = 20  2V = 40V (b) I o = 50  1mA = 50mA + V s = 2V V o = 20V s - (a) (b) 50I s I s = 1mA IoIo

Fall 2001Circuits I36 Example 8 - Compute the power that is supplied or absorbed by each element in the network below. Solution: P 36V = -(36V)(4A) = -144W (supplies 144W) P 1 = +(12V)(4A) = +48W (absorbs 48W) P 2 = +(24V)(2A) = +48W (absorbs 48W) P DS = -(4V)(2A) = -8W (supplies 8W) P 1 = +(28V)(2A) = +56W (absorbs 56W) Total = 0W 1I x I x = 4A 2A + 24V V V V -

Fall 2001Circuits I37 Example 9 - Determine the value of I o. Solution: P 2A = -(6V)(2A) = -12WP 1 = +(6V)(I o ) = 6I o W = ? P 2 = -(12V)(9A) = -108W P 3 = -(10V)(3A) = -30W P 4V = -(4V)(8A) = -32WP DS = +(16V)(11A) = 176W Since P Total = 0W = I o  6I o = 6W  I o = 1A - 6V I x I x = 2A 2A + 10V - + 4V - + 6V V + 9A IoIo 8A 3A 11A