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