CURRENT, VOLTAGE and RESISTANCE: Their Relationships

OHM’s LAW

A light bulb in a circuit can be lighted dim or very bright. For a

circuit like this, many factors must be considered to best describe

what might had happened on the light bulb.

Several answers will clearly describe the instance and will

surely focus on current, voltage & resistance.

CURRENT, VOLTAGE AND RESISTANCE… Their Relationships

When comparing the brightness of the bulb in each

case. There noticeable difference in each of the case.

This can lead to several conclusions and generalizations.

Case to investigate…

Case 1

Case 2

Case 3

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsAn important generalization

that can be made from the three (3) cases is that, the

number of dry cells increases, the brightness of the bulb also

increases. The dry cells in the setup

provided the emf that maintained the voltage across

the light bulb. With this, we can associate the increase in the number of dry cells used – increases the voltage across

the light bulb making it to glow brighter and brighter in each

of the cases.

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsThe activity simply shows the

concepts of the effect of voltage, electric current to

each other.“An increase in the voltage on the bulb results in an increase in electric current through the

bulb.”In 1837, Georg Simon Ohm discovered the relationship

between voltage and current. He summarized this idea in his formulated law – the “Ohm’s

Law”.

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsOhm’s law states that “the

ratio of voltage to current is constant known as resistance”.The law can be illustrated and

plotted into a graph of VOLTAGE-versus-CURRENT.V I (simply mean that V is

directly proportional to I)V

I

VI

--- = R (constant)

R (constant)

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsResistance is the property of the wire that is not affected by

voltage nor current through the wire.

The unit of resistance is equal to Volt per Ampere. The unit V/A is equivalent to the unit Ohm (represented by the

Greek letter omega, ) in the honor of Georg Simon Ohm.Ohm’s law is only valid for most metallic conductors. Conductors that behave

according to Ohm’s Law are called “ohmic conductors”.

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsLet’s apply Ohm’s Law to a practical example.A hot plate of an electric stove draws a current of 5A when plugged to a 220V source. What is the resistance of the hot plate?To compute…

I = 5A V = 220V R = ?R = V/I R = 220V / 5A = 44 V/A

R = 44 ohms or R = 44

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsOhm’s Law applied to another practical example.What is the current in the circuit which consists of a 12V energy source and a load of 9 ohms?

To compute…R = 9 V = 12V I = ?

I = V/R I = 12V / 9 = 44 V/

I = 1.33 Amperes or I = 1.33 A

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsOhm’s Law applied to another practical example.A resistor is connected to a 6-Volts battery. The ammeter in the circuit reads 2A. What is the value of the resistance?To compute…

V = 6V I = 2A R = ?R = V/I R = 6V / 2A = 3 V/A

R = 3 ohms orR = 3

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsOhm’s Law applied to another practical example.An electric iron rated 220V has a resistance of 60. Calculate the current through it when it is connected to a wall socket and switched on?To compute…R = 60 A V = 220V I

= ?I = V/R I = 220V / 60 = 3.67 V/A

I = 3.67 Amperes or I = 3.67 A

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsSolve the worded problem in ¼ by following the standard format. Required items: given, unknown,

formula & solving.A resistance of 18 ohms is across a 3-V battery. What is the current in the circuit?To compute…R = 18 V = 3V I = ?I = V/R I = 3V / 18

= 0.17 V/ I = 0.17 Amperes or I

= 0.17 A

CURRENT, VOLTAGE AND RESISTANCE… Their RelationshipsTo use Ohm’s Law, one must

remember the effect of each quantity present in the circuit. Another important item is the

formula/equation.The equation can be

associated to a triangular formula as used also in the format in the Newton’s 2nd

Law.And here it is…V = IRI = V/R R = V/I

VRI