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

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
DC Voltage Supply
Resistance


R = ρ L/A
ρ is the resistivity of
the material (units?)
Material
ρ (10-8 Ohm-Metres)
Silver
Copper
1.645
1.723
Gold
Aluminum
Tungsten
2.443
2.825
5.485
Nickel
Iron
Tantalum
Nichrome
Tin Oxide
Carbon
7.811
12.299
15.54
99.72
250
3500
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
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

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
Band 3 (special cases)

Gold = 0.1
 Red

Blue Gold = 2.6 Ohm
Silver = 0.01
 Red
Blue Silver = 0.26 Ohm
More Bands
Band 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%
Ohm’s Law
I = V/R
 V=IR
 R=V/I

Power

Power dissipated by charge flowing
through a resistor
P
= VI
 P = V2/R
 P = I 2R
Energy

Energy = Power x Time
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
Power Supply

Used very frequently in all devices.
 Transform
the AC supply into a lower voltage
 Rectify it (?)
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
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?

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?

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.

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.

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
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.
Resistance



The resistance seen
by the source
R=R1+R2
The two circuits on
the right are
equivalent
R1
R2
R1+R2
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
Power?

Power dissipated in each resistor
= V12/R1
 P1 = (V2/RT2)x R1
 P1

Total power = V2/RT = P1 + P2 + …
Kirchhoff’s Voltage Law

The algebraic sum of the potential rises
and drops around a closed loop is zero.
KVL


V + V1+V2 = 0
Can anyone prove
this mathematically?
R1
V1
V
R2
V2
Voltage Divider Rule

In a series circuit the voltage across the
resistive elements will divide as the
magnitude of the resistors
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.
Internal Resistances

Voltage and other sources have internal
resistances, and they should be counted
while solving circuits.
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
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
Current Divider Rule

For parallel elements of different value the
current will split with a ratio equal to the
inverse of their resistor value
Reduce and Return Approach

Applicable to all single source circuits.
Currents in Loops
(Remember KCL?)
i1
i2
i3
Ladder Networks

There are 2 approaches to solve this circuit
 Reduce
Resistances using series parallel analysis
 Calculate current using current loops
No - Load

When is “no load”
observed?
 When
large?
 Or when R2 is zero?
R1
R2
R2 is infinitely

When R2 is infinite!!!
Voltage Divider Supply
V1
R1

V2
R2

V3
R3
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.
Potentiometer
Voltage Sources
Two voltage sources of different ratings
may not be connected in parallel.
 Why?


Series operation, however, is permitted.
Current Sources

Two current sources of different rating may
not be connected in series.

Parallel operation, however, is permitted.