arrange the following in the order of current, starting with the lowest. (a) 7 coulombs in 3...

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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 Ohms (D) 2 Amps

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Page 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

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

Page 2: 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

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 1018 electrons/second 2 Amp = 12.484 x 1018 electrons/second In one hour - > 3600 x 12.484 x 1018 electrons Answer is 4.49 x 1022 electrons

Page 3: 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

DC Voltage Supply

Page 4: 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

Resistance

R = ρ L/A ρ is the resistivity of

the material (units?)

Page 5: 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

Material ρ (10-8 Ohm-Metres)

Silver 1.645

Copper 1.723

Gold 2.443

Aluminum 2.825

Tungsten 5.485

Nickel 7.811

Iron 12.299

Tantalum 15.54

Nichrome 99.72

Tin Oxide 250

Carbon 3500

Page 6: 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

American Wire Gage (AWG) sizes

AWG # Diameter (in) Ω /1000ft.0000 0.46 0.0490

000 0.409 0.0618

0 0.325 0.0983

1 0.289 0.1240

2 0.257 0.1563

4 0.204 0.2485

10 0.102 0.9989

14 0.0640 2.525

28 0.0126 64.90

Page 7: 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

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

Page 8: 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

Color Code (Band 1-3)

Color Value

Black 0

Brown 1

Red 2

Orange 3

Yellow 4

Green 5

Blue 6

Violet 7

Gray 8

White 9

Page 9: 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

Band 3 (special cases)

Gold = 0.1Red Blue Gold = 2.6 Ohm

Silver = 0.01Red Blue Silver = 0.26 Ohm

Page 10: 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

More BandsBand 4 Tolerance

Gold 5%

Silver 10%

None 20%

Band 5 Reliability (after 1000 Hrs of use)

Brown 1%

Red 0.1%

Orange 0.01%

Yellow 0.001%

Page 11: 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

Ohm’s Law

I = V/R V=IR R=V/I

Page 12: 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

Power

Power dissipated by charge flowing through a resistor P = VIP = V2/RP = I2R

Page 13: 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

Energy

Energy = Power x Time

Page 14: 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

Battery

Chemical Reactions to produce potential differenceAlkaline and lithium-iodine primary cellsLead Acid secondary cellNickel-Cadmium Secondary cellNickel-Hydrogen and Nickel-Metal Hydride

Secondary cells Solar Cells

Page 15: 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

Power Supply

Used very frequently in all devices. Transform the AC supply into a lower voltageRectify it (?)

Page 16: 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

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

Page 17: 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

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?

Page 18: 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

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?

Page 19: 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

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.

Page 20: 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

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.

Page 21: 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

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

Page 22: 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

Series Circuit

Two elements are in series ifThey have only one terminal in common. The common point in the two elements is not

connected to a third current carrying element.

Page 23: 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

Resistance

The resistance seen by the source

R=R1+R2

The two circuits on the right are equivalent

R1

R2

R1+R2

Page 24: 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

Voltage Drop?

The current through each resistor is calculated by the Ohm’s law=V1/R1

Where V1 is the voltage across the resistor.

=V/RT

Where RT is the total resistance in the circuit.

V1 = VxR1/RT

Page 25: 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

Power?

Power dissipated in each resistorP1 = V1

2/R1

P1 = (V2/RT2)x R1

Total power = V2/RT = P1 + P2 + …

Page 26: 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

Kirchhoff’s Voltage Law

The algebraic sum of the potential rises and drops around a closed loop is zero.

Page 27: 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

KVL

V + V1+V2 = 0 Can anyone prove

this mathematically?

R1

R2V

V1

V2

Page 28: 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

Voltage Divider Rule

In a series circuit the voltage across the resistive elements will divide as the magnitude of the resistors

Page 29: 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

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.

Page 30: 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

Internal Resistances

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

Page 31: 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

Resistance

R = ρ L/(A1+A2) Solving in terms of R1

and R2 gives 1/R = 1/R1 + 1/R2

The total value of the resistance is always smaller than the smallest resistance

Page 32: 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

Kirchhoff’s Current Law

KCL states that the algebraic sum of the currents entering and leaving a point or junction is zero.

i1+i2+i3+i4=0

i1

i2 i3 i4

Page 33: 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

Current Divider Rule

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

Page 34: 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

Reduce and Return Approach

Applicable to all single source circuits.

Page 35: 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

Currents in Loops (Remember KCL?)

i1i3i2

Page 36: 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

Ladder Networks

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

Page 37: 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

No - Load

When is “no load” observed? When R2 is infinitely

large? Or when R2 is zero?

When R2 is infinite!!!

R1

R2

Page 38: 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

Voltage Divider Supply

We cannot calculate V2 and V3 unless we know what load is connected to them

The less the load, the closer they are to 9V.

R1

R2

R3

V3

V2

V1

Page 39: 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

Potentiometer

Page 40: 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

Voltage Sources

Two voltage sources of different ratings may not be connected in parallel.

Why?

Series operation, however, is permitted.

Page 41: 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

Current Sources

Two current sources of different rating may not be connected in series.

Parallel operation, however, is permitted.