beee1ation (2).pptx

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Ohm’s Law Relationship between current and voltage within a circuit element. The voltage across an element is directly proportional to the current flowing through it v α i • Thus::v=iR and R=v/i – Where: » R is called resistor . » Has the ability to resist the flow of electric current. » Measured in Ohms (Ω)

Transcript of beee1ation (2).pptx

Ohms Law

Ohms LawRelationship between current and voltage within a circuit element.

The voltage across an element is directly proportional to the current flowing through it v i

Thus::v=iR and R=v/iWhere:R is called resistor.Has the ability to resist the flow of electric current.Measured in Ohms ()Ohms LawConductance (G) Unit mho or Siemens (S).

Reciprocal of resistance R

G = 1 / R Has the ability to conduct electric currentOhms LawPower:

P = iv i ( i R ) = i2R watts (v/R) v = v2/R watts

R and G are positive quantities, thus power is always positive.

R absorbs power from the circuit Passive element.Ohms LawExample 1:Determine voltage (v), conductance (G) and power (p) from the figure below.

Ohms LawExample 2:Calculate current i in figure below when the switch is in position 1.Find the current when the switch is in position 2.

DKS1113 Electric CircuitsNodes, Branches & LoopsExample 4:Determine how many branches and nodes for the following circuit.

Kirchoffs Laws

DKS1113 Electric CircuitsKirchoffs LawsExample 6:Current in a closed boundary

Kirchoffs LawsExample 11:Use KVL to obtain v1, and v2.

Series and parallel resistors Kirchhoffs Rules for network problems Electrical meters and household circuits R-C circuits

DC CIRCUITS: Chapter 26C 2012 J. BeckerDC CIRCUITS

C 1998 McDermott, et al., Prentice HallCalculate the total resistance of each box if each bulb has a resistance of 1000 Ohms (W).500, 1000, 2000, 670 Ohms

Method of simplifying the circuit in (a) below to get the equivalent resistance. We can then calculate the power P = I2 R dissipated in each resistor.27-2 Simplifying series and parallel resistors (Example 27-1)

Exercise 26.13: Four equivalent light bulbs R1 = R2 = R3 = R4 = 4.50 W, emf = 9.00 Volts.Find current and power in each light bulb. Which bulb is brightest?Later, if bulb #4 is removed which bulbs get brighter? Dimmer? REDRAW THE CIRCUITItotal=I1=1.5A,, I2=I3=I4+0.5A., P1=10.1W, P2=P3=P4=1.125W., V2=V3=V4=2.25V On the course website, see old Test #1 from previous semesters for typical DC circuit problems

AND other typical problems on tests

Seewww.physics.sjsu.edu/Becker/physics51Figure 26.11

Typo on top of page 890 (12th Edition), 2nd line from top of page:right side should read bottom(See Figure 26.11 which is rotated 90o from figure in the 11th Edition.)

DISCHARGING: CHARGING: An RC circuit that can be used to charge and discharge a capacitor (through a resistor).27-34 RC circuit

CHARGING A CAPACITOR:current vs timeCHARGING A CAPACITOR:charge vs time

DISCHARGING A CAPACITOR:current vs time

DISCHARGING A CAPACITOR:charge vs time

ooA battery, a capacitor, and a resistor are connected in series. Which of the following affect(s) the maximum charge stored on the capacitor?Q26.26A. the emf e of the batteryB. the capacitance C of the capacitorC. the resistance R of the resistorD. both e and CE. all three of e, C, and R

Answer: D

House wiring circuits27-20 House wiring system

Hand drill circuit with ground wire for safety27-21 Hand drill circuit with ground wire

see www.physics.sjsu.edu/Becker/physics51ReviewC 2012 J. F. BeckerOVERVIEWConductance (G)Series conductance:1/Geq = 1/G1 +1/G2+

Parallel conductance:Geq = G1 +G2+

DKS1113 Electric CircuitsVoltage and Current DivisionCalculate v1, i1, v2 and i2.

Voltage and Current DivisionDetermine i1 through i4.

Voltage and Current DivisionExample 15:Determine v and i.Answer v = 3v, I = 6 A.

DKS1113 Electric CircuitsVoltage and Current DivisionExample 16:Determine I1 and Vs if the current through 3 resistor = 2A.

DKS1113 Electric CircuitsVoltage and Current DivisionExample 17:Determine Rab.

DKS1113 Electric CircuitsVoltage and Current DivisionExample 18:Determine vx and power absorbed by the 12 resistor.Answer v = 2v, p = 1.92w.

DKS1113 Electric CircuitsWye-Delta TransformationsGiven the circuit, how to combine R1 through R6?Resistors are neither in series nor parallel

Use wye-delta transformations

Wye-Delta Transformations Y networkT network

DKS1113 Electric CircuitsWye-Delta Transformations network network

DKS1113 Electric CircuitsWye-Delta TransformationsDelta () to wye (y) conversion.

DKS1113 Electric CircuitsWye-Delta TransformationsExample 19:Transform the circuit from to y.Answer R1=18, R2=6, R3=3.

Wye-Delta TransformationsExample 20:Determine Rab.Answer Rab=142.32.

DKS1113 Electric CircuitsWye-Delta TransformationsExample 21:Determine Io.

DKS1113 Electric CircuitsBasic Electrical Principles

Conductors - keep loose grip on their electrons and allow electrons to move freely. Metals are usually good conductors.

Insulators - keep close hold of their electrons and do not allow free movement of electrons. Glass, wood, plastic, mica, fiberglass and air are good insulators.

Electromotive Force (EMF) is the force that moves electrons through conductors. Its unit of measure is the Volt. Think of it as pressure.

Voltage Source - has two terminals (+ and -). Some examples are car batteries (12 volts DC), D cell batteries (1.5 volts DC) and a wall socket (120 volts AC).

Current - is the flow of electrons. It is measured in amperes.

Resistance (ohms, ) is the ability to oppose an electrical current.