Current Electricity

Current Electricty

• Unlike Static electricity which does not flow, Current electricity “flows” through a circuit.

• The electrons flow from the negative pole to the positive pole. There has to be something that forces the electrons back to the negative pole so that they can flow again. This is what a battery or generator does.

Current Electricity

Electrons are negatively charged and flow from the negativepole to the positive. A battery (or generator) pushes the electronsback to the negative pole so that they can flow to the positive. The greater the difference between these two poles, the greater the Electrical Potential Difference.

Current Electricity

• Voltage is the potential difference between the Positive and Negative Poles.

• The greater the difference, the greater the voltage and the more work that it can do.

• As electricity flows through a circuit, the voltage “drops” at each resistor until it equals zero at the end of the circuit.

• The battery (or generator) raises the voltage to the original volts and the electricity flows through the circuit again.

Current Electricity

Current Electricity- +

1.5 V

V = 1.5 V V = 0 V

Current Electricity

• Amperage is the “rate of flow” of the electrons• The higher the amperage of a circuit, the faster the

electrons are “flowing”• 1 Amp = 1 Coulomb/sec.• The symbol for amperage is I• The unit for amperage is amps.• There are two factors that determine the amperage

of a circuit:– The resistance to the flow of electrons.– The voltage, or the electric potential.

Current Electricity

• Resistance is as the name implies, something that resists the “flow” of electrons

• Anything that uses electricity is a resistor

• Resistance is measured in ohms ()

Ohm’s Law

• Relates the three components of a circuit

• Amperage = Voltage/ Resistance

• I = V/R

• or V = I R

Simple Circuits(only one path)

A amp meter

Simple Circuits(only one path)

A amp meter

What is the current at A if there is 12 Volts and the lightbulb has 0.5 of resistance?I = V/R , I = 12 V/0.5 , I = 24 Amps

Simple Circuit

What is the resistance of the light bulb (R1) if there is 120 Voltsand the current is 15 Amps?R = V/I R = 120V/15A R = 8

R1

120V

15 Amps

Series Circuits

In a Series Circuit, there are more than one resistor, but thereis only one path. for the electrons to follow.

Series Circuits

Before you can solve for voltage or current, you need to know the total resistance of the CircuitRT = R1+R2 + R3 ….

I = V/ RT

Series Circuits15 10 15

24 V

Find RT

RT= 15 + 10 + 15

RT = 40

I = V/RI = 24V/40 I = 0.6 A

A

Parallel Circuits

In a parallel circuit, there are more than one pathfor the electrons to follow. You have to calculatewhat the equivalent resistance would be. In otherwords, what one resistor would have the same valueas the three resistors in the circuit.

Parallel Circuits

Since there are more than one path, the current actualincreases as more paths are added. The total resistanceis an inverse relationship. As resistors are added,the total resistance decreases.

Parallel Circuits

Equivalent Resistance =1 = 1 + 1 + 1 ……RT R1 R2 R3

1 = 1 + 1 + 1 = 5RT 30 15 15 30Equivalent R = 30 = 6 5

R1 = 30 R2 = 15 R3 = 15

120V

Parallel Circuits6

120 V A

The 3 resistors are replaced by the equivalentresistor of 6 . Now you can calculate thecurrent in the circuit.

Parallel Circuits

Equivalent Resistance = 6

I = V/RT

I = 120 V / 6 I = 20 Amps

R1 = 30 R2 = 15 R3 = 15

120 V A

Combined Circuits

R5 75

A

R5 75

In a combined circuit problem, you first have to find all of the equivalentresistances in the parallel portions.

1 = 1 + 1 = 7 500 = 71.42 R1+2 100 250 500 7

1 = 1 + 1 = 4 + 7 1400 = 127.27 R3 +4 350 200 1400 1400 11

A

R5 75 Now that the parallel portions have been replaced by their equivalents,you treat the circuit as a series circuit and find the total resistance.

RT = 71.429 + 127.27 + 75 = 273.70

I = V/RI = 24 V / 273.70 I = 0.088 AmpsThis is the current in the entire circuit, not the current in the parallel portions.

A

273.70

24 V

Now that the parallel portions have been replaced by their equivalents,

you treat the circuit as a series circuit and find the total resistance.

RT = 71.429 + 127.27 + 75 = 273.70

I = V/R

I = 24 V / 273.70

I = 0.088 Amps

This is the current in the entire circuit, not the current in the parallel portions.

A

Finding Voltage Drops

V1

V2

R5 75

V3

Find the voltage drops at each meter.

Voltage Drops

To find the voltage drops in a parallel portion, you use theEquivalent resistance of that portion and the current for theentire circuit. V1 = I x R1+2 = (0.088 A)(71.42 V) = 6.28 V

V2 = I x R3 + 4 = (0.088 A)(127.27 V) = 11.20 V

V3 = I x R5 = (0.088 A)(75 ) = 6.6 V

All of the voltage drops should equal the 24.08 Vtotal voltage of the circuit

Current in Parallel Portions

R5 75

A1

A2 A3

A4 A5

Current in Parallel PortionsThe current in the parallel portions is not the same asin the main circuit. It gets divided depending on the actualresistor ( not the equivalent). You need to know the voltagedrop at each resistor before you can calculate the current. A1 = is the current in the main portion = 0.088 AmpsA2 = V1 / R1 = 6.28 V/ 100 = 0.0628 AmpsA3 = V1/R2 = 6.28 V / 250 = 0.0251 Amps(It should equal the total amps) 0.08792 Amps

A4 = V2/ R3 = 11.2 V/ 350 = 0.032 AmpsA5 = V2/ R4 = 11.2 V/ 200 = 0.056 Amps(It should equal the total amps) 0.088 Amps