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Arrange the following in the order of current, starting with the lowest. (A) 7 coulombs in 3 seconds. (B) 10 Volts across 5 Ohms (C) 15 Volts across 7.

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Presentation on theme: "Arrange the following in the order of current, starting with the lowest. (A) 7 coulombs in 3 seconds. (B) 10 Volts across 5 Ohms (C) 15 Volts across 7."— Presentation transcript:

1 Arrange the following in the order of current, starting with the lowest. (A) 7 coulombs in 3 seconds. (B) 10 Volts across 5 Ohms (C) 15 Volts across 7 Ohms (D) 2 Amps

2 Question If a laptop needs constantly needs 2 Amps current from a battery, how many electrons are drained from the battery in one hour? 1 Amp = 6.242 x 10 18 electrons/second 2 Amp = 12.484 x 10 18 electrons/second In one hour - > 3600 x 12.484 x 10 18 electrons Answer is 4.49 x 10 22 electrons

3 DC Voltage Supply

4 Resistance R = ρ L/A ρ is the resistivity of the material (units?)

5 Material ρ (10 -8 Ohm-Metres) Silver1.645 Copper1.723 Gold2.443 Aluminum2.825 Tungsten5.485 Nickel7.811 Iron12.299 Tantalum15.54 Nichrome99.72 Tin Oxide250 Carbon3500

6 American Wire Gage (AWG) sizes AWG #Diameter (in)Ω /1000ft. 00000.460.0490 0000.4090.0618 00.3250.0983 10.2890.1240 20.2570.1563 40.2040.2485 100.1020.9989 140.06402.525 280.012664.90

7 Color Coding 5 Bands of code (3 are mandatory) Bands 1 - 3  the value of the resistor Band 4  the range (tolerance) Band 5  the reliability

8 Color Code (Band 1-3) ColorValue Black0 Brown1 Red2 Orange3 Yellow4 Green5 Blue6 Violet7 Gray8 White9

9 Band 3 (special cases) Gold = 0.1  Red Blue Gold = 2.6 Ohm Silver = 0.01  Red Blue Silver = 0.26 Ohm

10 More Bands Band 4Tolerance Gold5% Silver10% None20% Band 5Reliability (after 1000 Hrs of use) Brown1% Red0.1% Orange0.01% Yellow0.001%

11 Ohm’s Law I = V/R V=IR R=V/I

12 Power Power dissipated by charge flowing through a resistor  P = VI  P = V 2 /R  P = I 2 R

13 Energy Energy = Power x Time

14 Battery Chemical Reactions to produce potential difference  Alkaline and lithium-iodine primary cells  Lead Acid secondary cell  Nickel-Cadmium Secondary cell  Nickel-Hydrogen and Nickel-Metal Hydride Secondary cells Solar Cells

15 Power Supply Used very frequently in all devices.  Transform the AC supply into a lower voltage  Rectify it (?)

16 Current Sources Supplies a fixed amount of current  It is the dual of the battery In a battery voltage is constant, but current drains out

17 Ammeters Device to measure current The wire in which current is to be measured is broken up, and are joined via an ammeter. What should be the resistance of the ammeter?

18 Voltmeters Devices to measure voltage Connected in a parallel fashion across the device where there is a need to measure potential difference. What should be the resistance of a voltmeter?

19 Ohmmeters Used to measure the resistance of a device. Connected across the two pins of a resistor Also used to check the continuity of networks.

20 Wattmeters Used to measure the dissipation of power in a circuit element. Includes both an ammeter and a voltmeter. Displays the multiplication of both measurements.

21 Fuses and Circuit Breakers The power supply to the homes is not ideal. If it goes above a specified level, it can burn the devices. May result in Fire or Smoke. Fuse wires melt if they experience a large current. In a breaker, a large current results in a large enough strength in an inbuilt electromagnet to draw the switch open

22 Series Circuit Two elements are in series if  They have only one terminal in common.  The common point in the two elements is not connected to a third current carrying element.

23 Resistance The resistance seen by the source R=R 1 +R 2 The two circuits on the right are equivalent R1R1 R2R2 R 1 +R 2

24 Voltage Drop? The current through each resistor is calculated by the Ohm’s law  =V 1 /R 1  Where V 1 is the voltage across the resistor.  =V/R T  Where R T is the total resistance in the circuit. V 1 = VxR 1 /R T

25 Power? Power dissipated in each resistor  P 1 = V 1 2 /R 1  P 1 = (V 2 /R T 2 )x R 1 Total power = V 2 /R T = P 1 + P 2 + …

26 Kirchhoff’s Voltage Law The algebraic sum of the potential rises and drops around a closed loop is zero.

27 KVL V + V 1 +V 2 = 0 Can anyone prove this mathematically? R1R1 R2R2 V V1V1 V2V2

28 Voltage Divider Rule In a series circuit the voltage across the resistive elements will divide as the magnitude of the resistors

29 Ground Terminal This is not a loop.  Or is it? Ground terminal means that the two points are both connected to ground and are at a zero potential.  So this is a loop.

30 Internal Resistances Voltage and other sources have internal resistances, and they should be counted while solving circuits.

31 Resistance R = ρ L/(A 1 +A 2 ) Solving in terms of R1 and R2 gives  1/R = 1/R 1 + 1/R 2 The total value of the resistance is always smaller than the smallest resistance

32 Kirchhoff’s Current Law KCL states that the algebraic sum of the currents entering and leaving a point or junction is zero. i 1 +i 2 +i 3 +i 4 =0 i1i1 i2i2 i3i3 i4i4

33 Current Divider Rule For parallel elements of different value the current will split with a ratio equal to the inverse of their resistor value

34 Reduce and Return Approach Applicable to all single source circuits.

35 Currents in Loops (Remember KCL?) i1i1 i3i3 i2i2

36 Ladder Networks There are 2 approaches to solve this circuit  Reduce Resistances using series parallel analysis  Calculate current using current loops

37 No - Load When is “no load” observed?  When R 2 is infinitely large?  Or when R 2 is zero? When R 2 is infinite!!! R1R1 R2R2

38 Voltage Divider Supply We cannot calculate V 2 and V 3 unless we know what load is connected to them The less the load, the closer they are to 9V. R1R1 R2R2 R3R3 V3V3 V2V2 V1V1

39 Potentiometer

40 Voltage Sources Two voltage sources of different ratings may not be connected in parallel. Why? Series operation, however, is permitted.

41 Current Sources Two current sources of different rating may not be connected in series. Parallel operation, however, is permitted.

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