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Ohm’s Law & Circuits
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A. Voltage- electrical potential; an electrical pressure created by the buildup of charge; causes charged particles to move
B. Volt- unit of voltage; Symbol= VC. Electromotive force- a historical term used to describe
voltage; Symbol= E (No longer relevant, the definition of force is something that causes a mass to accelerate, and voltage or EMF does not fit that definition). E is now commonly used as a symbol for electric field strength.
D. Current- the flow or movement of electronsE. Ampere- unit of current; Symbol= I F. Resistance- opposition to current flowG. Ohm- unit of resistance; Symbol= Ω (Greek symbol Omega)
H. Energy- the fundamental ability to do workI. Joule- unit of energy; Symbol= JJ. Electrical Power- the rate of electrical energy used in a circuit; calculated by multiplying current times voltage,
or P = V • IK. Watt- unit of measurement for power; a watt is one
joule per second (J/s); Symbol= WL. Ohm’s Law- a formula describing the mathematical
relationship between voltage, current, and resistance; one of the most commonly used equations in all of science
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M. Directly proportional- having a constant ratio; a situation where one variable moves in the same direction as another variable when other conditions are constant
M. Inversely proportional- having a constant but inverse ratio; a situation where one variable moves in the opposite direction from another variable when other conditions remain constant
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• Example- current doubles when voltage is doubled if resistance is held constant; thus, voltage and current are directly proportional
• Example- with a constant voltage, current decreases when resistance increases; thus, current and resistance are inversely proportional
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• I – Electrical current in amperes• R – Resistance in ohms• V – Represents voltage in volts• A – Represents amperes• Ω – Represents ohms• E – Electromotive force (emf) in volts,
sometimes used as an alternate symbol for voltage
Power is an indication of how much work (the conversion of energy from one form to another) can be done in a specific amount of time; that is, a rate of doing work.
Power can be delivered or absorbed as defined by the polarity of the voltage and the direction of the current.
tW
P
second / joule 1 (W)Watt 1
Energy (W) lost or gained by any system is determined by:
W = Pt Since power is measured in watts (or
joules per second) and time in seconds, the unit of energy is the wattsecond (Ws) or joule (J)
The watt-second is too small a quantity for most practical purposes, so the watt-hour (Wh) and kilowatt-hour (kWh) are defined as follows:
The killowatt-hour meter is an instrument used for measuring the energy supplied to a residential or commercial user of electricity.
1000(h) time (W) power
(kWh)Energy
(h) time (W) power (Wh)Energy
Efficiency () of a system is determined by the following equation:
= Po / Pi
Where: = efficiency (decimal number) Po = power output Pi = power input
The basic components of a generating (voltage) system are depicted below, each component has an associated efficiency, resulting in a loss of power through each stage.
Insert Fig 4.19Insert Fig 4.19
Insert Table 4.1Insert Table 4.1
Basic Laws of CircuitsOhm’s Law:
The voltage across a resistor is directly proportional to the currentmoving through the resistor.
+ _v (t)i(t)
Rv (t) = R i(t)
+_ v (t)i(t)
Rv (t) = R i(t)_
(2.1)
(2.2)
Basic Laws of CircuitsOhm’s Law:
Directly proportional means a straight line relationship.
v(t)
i(t)
R
The resistor is a model and will not produce a straight linefor all conditions of operation.
v(t) = Ri(t)
Basic Laws of CircuitsOhm’s Law:About Resistors:The unit of resistance is ohms( ). A mathematical expression for resistance is
lRA
2
: ( )
: ( )
:
l Thelengthof theconductor meters
A Thecross sectional area meters
The resistivity m
(2.3)
Basic Laws of CircuitsOhm’s Law:About Resistors:
We remember that resistance has units of ohms. The reciprocal ofresistance is conductance. At one time, conductance commonly had unitsof mhos (resistance spelled backwards).
In recent years the units of conductance has been established as seimans (S).
Thus, we express the relationship between conductance and resistance as
1GR
(2.4)
We will see later than when resistors are in parallel, it is convenientto use Equation (2.4) to calculate the equivalent resistance.
(S)
Basic Laws of CircuitsOhm’s Law:Ohm’s Law: Example 2.1.
Consider the following circuit.
+
_1 1 5 V R M S
(a c )R(1 0 0 W a tt lig ht bu lb)
V
Determine the resistance of the 100 Watt bulb.
2
22
2 115 132.25100
VP VI I RR
VR ohmsP
(2.5)
A suggested assignment is to measure the resistance of a 100 watt lightbulb with an ohmmeter. Debate the two answers.
Resistivity is a material property• Dependent on the number of free or mobile
charges (usually electrons) in the material. In a metal, this is the number of electrons from
the outer shell that are ionized and become part of the ‘sea of electrons’
• Dependent on the mobility of the charges Mobility is related to the velocity of the charges. It is a function of the material, the frequency and
magnitude of the voltage applied to make the charges move, and temperature.
Material Resistivity (-cm) UsageSilver 1.64x10-8 ConductorCopper 1.72x10-8 ConductorAluminum 2.8x10-8 ConductorGold 2.45x10-8 ConductorCarbon (Graphite) 4x10-5 ConductorGermanium 0.47 SemiconductorSilicon 640 SemiconductorPaper 1010 InsulatorMica 5x1011 InsulatorGlass 1012 InsulatorTeflon 3x1012 Insulator
mA Milliamp = 0.001 amps
Kilo ohm = 1000 ohmsKΩ
Resistance takes into account the physical dimensions of the material
where:L is the length along which the carriers are moving
A is the cross sectional areathat the free charges movethrough.
ALR
Voltage drop across a resistor is proportional to the current flowing through the resistor
Units: V = Awhere A = C/s
iRv
Insert Fig 4.8Insert Fig 4.8
If the resistor is a perfect conductor (or a short circuit)R = 0 ,
then
v = iR = 0 V
no matter how much current is flowing through the resistor
If the resistor is a perfect insulator, R = ∞
then
no matter how much voltage is applied to (or dropped across) the resistor.
A
Conductance is the reciprocal of resistance
G = R-1 = i/vUnit for conductance is S (siemens) or
(mhos)
G = A/L where is conductivity,
which is the inverse of resistivity,
p = iv = i(iR) = i2R
p = iv = (v/R)v = v2/R
p = iv = i(i/G) = i2/G
p = iv = (vG)v = v2G
Since R and G are always real positive numbers• Power dissipated by a resistor is always
positive The power consumed by the resistor is
not linear with respect to either the current flowing through the resistor or the voltage dropped across the resistor• This power is released as heat. Thus, resistors
get hot as they absorb power (or dissipate power) from the circuit.
There is no power dissipated in a short circuit.
There is no power dissipated in an open circuit.
W0)0()V0(v 22 Rpsc
W0)()A0(i 22 Rpoc
The path that the current follows is called an electric circuit.
All electric circuits consist of a voltage source, a load, and a conductor
The voltage source establishes a difference of potential that forces the current to flow.
A series circuit offers a single path for current flow.
A parallel circuit offers more than one path for current flow.
A series-parallel circuit is a combination of a series circuit and a parallel circuit.
How much current flows in the circuit shown in Figure 1?
Figure 1
Given: IT = ?
ET = 12 volts RT= 1000 ohms
IT = ET/ RT
IT = 12 /1000IT = 0.012 amp or 12 milliamps
In the circuit shown in Figure 2, how much voltage is required to produce 20 milliamps of current flow
Figure 2
Given:
IT = 20 mA = = 0.02 amp
ET = ? RT= 1.2 kΩ = ohms
In the circuit shown in Figure 5-10, how much voltage is required to produce 20 milliamps of current flow?
Solution:
IT = ET / RT
0.02 = ET / 1200ET = 24 volts
What resistance value is needed for the circuit shown in Figure 5-11 to draw 2 amperes of current?
Given : IT = 2 ampsET = 120 volts RT = ?
Solution: IT = ET / RT2 = 120 / RT60 ohms = RT
What is the total current flow in the circuit?
Given:IT = ? ET = 12 volts RT =? R1 = 560 ohms R2 = 680 ohms R3 = 1 kΩ= 1000 ohmsSolution: First solve for the total resistance of the circuit: RT = R1 + R2 + R3 RT = 560 + 680 + 1000 = 2240 ohmsRT = 2240 ohms
Draw an equivalent circuit. See Figure. Now solve for the total current flow:
IT = ET / RT IT = 12 / 2240 IT = 0.0054 amp or 5.4 milliamp
How much voltage is dropped across resistor R2 in the circuit
Given:IT =? ET = 48 volts RT =? R1 = 1.2 k Ω = 1200 ohms R2 = 3.9 k Ω = 3900 ohms R3 = 5.6 k Ω = 5600 ohmsSolution: First solve for the total circuit resistance:
RT = R1 + R2 + R3 = 1200 + 3900 + 5600 RT = 10,700 ohms Draw the equivalent circuit. See Figure. Solve for the total current in the circuit: IT = ET / RT IT = 48 / 10,700 IT = 0.0045 amp or 4.5 milliamps
Remember, in a series circuit, the same current flows throughout the circuit. Therefore, IR = IT
IR2 = ER2 / R2
0.0045 = ER / 3900 ER2 = 17.55 volts
What is the value of R2in the circuit shown in Figure
What is the value of R2 in the circuit shown in Figure?
First solve for the current that flows through R1 and R3. Because the voltage is the same in each branch of a parallel circuit, each branch voltage is equal to the source voltage of 120 volts.
Given: IR1
ER1 = 120 voltsR1 = 1000 ohms
Solution:IR1 = ER1 / R1IR1 = 120 / 1000IR1 = 0.12 amp
Given: IR3
ER3 = 120 voltsR3 = 5600 ohmsSolution:
IR3 = ER3 / R3IR3 = 120 / 5600IR3 = 0.021 amp
Circuit, the total current is equal to the sum of the currents in the branch currents.
Given:IT = 0.200 amp IR1 = 0.120 amp IR2=? IR3= 0.021 ampSolution:IT = IR1 + IR2 + IR3
0.200 = 0.120 + IR2 + 0.021 0.200 = 0.141 + IR2 0.200 - 0.141 = IR2
Resistor R2 can now be determined using Ohm's law.
Given:IR2 = 0.059 ampER2 = 120 volts R2 = ?IR2 = ER2 / R2.Solution:0.059 = 120 / R2 R2 = 2033.9 ohms
What is the current through R3 in the circuit shown in Figure
First determine the equivalent resistance (RA) for resistors R1 and R2.
Given:RA =? R1 = 1000 ohms R2 = 2000 ohmsSolution:1 / RA = 1/R1 + 1/R2 (adding fractions requires a common denominator)
1 / RA = 1/1000 + 1/20001 /RA = 666.67 ohms
Then determine the equivalent resistance (RB) for resistors R4, R5, and R6• First, find the total series resistance (Rs) for resistors R5 and R6.
Given: Rs =? R5 = 1500 ohms R6= 3300 ohmsSolution: Rs = R5 + R6 Rs = 1500 + 3300 Rs = 4800 ohms
Given:RB =? R4 = 4700 ohms Rs = 4800 ohmsSolution: 1 / RB = 1/R4 + 1/Rs1 / RB = 1/4700 + 1/4800RB = 2375.30 ohms
Redraw the equivalent circuit substituting RA and RB , and find the total series resistance of the equivalent circuit. See Figure
Given:RT =? RA = 666.67 ohms R3 = 5600 ohms RB = 2375.30 ohms
Solution:RT = RA + R3 + RB RT = 666.67 + 5600 + 2375.30 RT = 8641.97 ohmsNow solve for the total current through the equivalent
circuit using Ohm's law.Given:IT =? ET = 120 volts RT = 8641.97 ohmsSolution:IT = ET / RTIT = 120 / 8641.97 IT = 0.0139 amp or 13.9 milliamps
In a series circuit, the same current flows throughout the circuit. Therefore, the current flowing through R3 is equal to the total current in the circuit.
IR3 = ITIR3 = 13.9 milliamps
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