Phys 2180 Lecture (5)

Current and resistance and Direct current circuits

• INTRODUCTIONIn the past chapters we have been discussing interactions of electric charges “at rest” (electrostatic).

Now we are ready to study charges “in motion”.

An electric current consists of motion of charges charges from one region to another.

When this motion of charges takes place within a conductor that forms a closed path, the path is called an electric circuit.

ELECTRIC CURRENTS

THE ELECTRIC CURRENT

We can also define current through the area as the net charge flowing through the area per unit time.

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CURRENT (Electric Current) It is the rate of flow of electric charge through a cross-

sectional area.

Electric CurrentThe electric current is the amount of charge per unit time that passes through a surface that is perpendicular to the motion of the charges.

The SI unit of electric current is the ampere (A), after the French mathematician André Ampére (1775-1836). 1 A = 1 C/s. Ampere is a large unit for current. In practice milliampere (mA) and microampere (μA) are used.

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OHM’S LAW:RESISTANCE AND RESISTORS

• RESISTANCE– The proportionality of J to E for a metallic

conductor at constant temperature was discovered by G.S. Ohm.

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For Ohmic materials (those that obey Ohm’s law), the potential V is proportional to the current I.

The behavior will always trace a linear relationship.

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OHM’S LAW:RESISTANCE AND RESISTORS

• RESISTOR

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Current I enters a resistor R as shown. (a) Is the potential higher at point A or at point B? (b) Is the current greater at point A or at point B?

ELECTRIC POWER

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Power dissipated in a conductor

Power dissipated in a resistor

A CIRCUIT is a closed conducting path current flow all the way around.

• The POWER is the work done per unit time or the time rate of energy transfer

EMF AND TERMINAL VOLTAGE

Ideal Emf SourceReal Battery

Electromotive Force (emf) Source It is a device that supplies electrical

energy to maintain a steady current in a circuit.

It is the voltage generated by a battery.

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RESISTORS INSERIES AND PARALLEL

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RESISTORS IN SERIES The magnitude of the charge is constant. Therefore, the

flow of charge, current I is also constant. The potential of the individual resistors are in general

different.

The equivalent resistance of resistors in series equals the sum of their individual resistances.

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Totalseries

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As the current goes through the circuit, the charges must USE ENERGY to get through the resistor. So each individual resistor will get its own individual potential voltage). We call this voltage drop.

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;Note: They may use the terms “effective” or “equivalent” to mean TOTAL!

Series Resistors and Voltage Division (1)

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Parallel Resistors and Current Division (1)Parallel Resistors and Current Division (1)• Parallel wiring means that the devices are connected in such a way

that the same voltage is applied across each device.

• Multiple paths are present.

• When two resistors are connected in parallel, each receives current from the battery as if the other was not present.

• Therefore the two resistors connected in parallel draw more current than does either resistor alone.

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RESISTORS INSERIES AND PARALLEL

Combinations of Resistors The 8.0- and 4.0-

resistors are in series and can be replaced with their equivalent, 12.0

The 6.0- and 3.0- resistors are in parallel and can be replaced with their equivalent, 2.0

These equivalent resistances are in series and can be replaced with their equivalent resistance, 14.0

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QuestionCalculate the total resistance of a 4 and 6 ohm resistor connected (a) in series, (b) in parallel.

(a) series

RT = R1 + R2

= 4 Ω + 6 Ω= 10 Ω

(b) parallel

1 / RT = 1 / R1 + 1 / R2

= 1 / (4 Ω) + 1 / (6 Ω)= 0.2500 + 0.1666= 0.4166

= 1 / RT !!!!

and so RT = 1 / 0.4166= 2.4 Ω 14

Calculate the total resistance of the two circuits shown below:

Calculate the parallel section first1 / R1+2 = 1 / R1 + 1 / R2

= 1 / (2 Ω) + 1 / (5 Ω)= 0.5000 + 0.2000= 0.7000R1+2 = 1.429 Ω

Add in series resistanceRT = 5.429 Ω

= 5.43 Ω (to 3sf)

4 Ω

2 Ω

5 Ω

1.

12 Ω

8 Ω5 Ω

2.

Calculate the series section first5 Ω + 8 Ω = 13 Ω

Calculate 13 Ω in parallel with 12 Ω 1 / RT = 1 / R1 + 1 / R2

= 1 / (13 Ω) + 1 / (12 Ω)= 0.07692 + 0.08333= 0.16025RT = 6.2402 Ω= 6.24 Ω (to 3sf) 15

The heating effect of an electric current• When an electric current flows through an electrical

conductor the resistance of the conductor causes the conductor to be heated.

• This effect is used in the heating elements of various devices like those shown below:

Heating effect of resistance Phet

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Electric Power,P.

time

EnergyP

Since the electrical energy is charge times voltage (QV), the above equation becomes,

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Since the current is charge flow per unit time (Q/t), the above equation becomes,

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Since V = IR, the above equation can also be written as,

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VRIIVP SI Unit of Power: watt(W)

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Power and resistanceRevision of previous work

When a potential difference of V causes an electric current I to flow through a device the electrical energy converted to other forms in time t is given by:

E = I V t

but: power = energy / time

Therefore electrical power, P is given by:

P = I V

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The definition of resistance: R = V / I

rearranged gives: V = I R

substituting this into P = I V gives:

P = I 2 R

Also from: R = V / I

I = V / R

substituting this into P = I V gives:

P = V 2 / R

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Question 1Calculate the power of a kettle’s heating element of resistance 18Ω when draws a current of 13A from the mains supply.

P = I 2 R

= (13A)2 x 18Ω

= 169 x 18

= 3042W

or = 3.04 kW

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Question 2Calculate the current drawn by the heating element of an electric iron of resistance 36Ω and power 1.5kW.

P = I 2 R gives:

I 2 = P / R

= 1500W / 36 Ω

= 41.67

= I 2 !!!!

therefore I = √ ( 41.67)

= 6.45 A

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Emf and internal resistance Emf, electromotive force (ε):

The electrical energy given per unit charge by the power supply.

Internal resistance (r):

The resistance of a power supply, also known as source resistance.

It is defined as the loss of potential difference per unit current in the source when current passes through the source.

ε = E

Q

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Equation of a complete circuitThe total emf in a complete circuit is equal to the total pds.

Σ (emfs) = Σ (pds)

For the case opposite:

ε = I R + I r

or

ε = I ( R + r )

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Question 2Calculate the current drawn from a battery of emf 1.5V whose terminal pd falls by 0.2V when connected to a load resistance of 8Ω.

ε = I R + I r

where I r = lost volts = 0.2V

1.5 V = (I x 8 Ω) + 0.2V

1.5 – 0.2 = (I x 8)

1.3 = (I x 8)

I = 1.3 / 8

current drawn = 0.163 A

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Question 3Calculate the terminal pd across a power supply of emf 2V, internal resistance 0.5Ω when it is connected to a load resistance of 4Ω.

ε = I R + I r

where I R = terminal pd

2 V = (I x 4 Ω) + (I x 0.5 Ω )

2 = (I x 4.5)

I = 2 / 4.5

= 0.444 A

The terminal pd = I R

= 0.444 x 4

terminal pd = 1.78 V

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Single cell circuit rules1. Current drawn from the cell: I = cell emf

total circuit resistance

2. PD across resistors in SERIES with the cell: V = cell current x resistance of each resistor

3. Current through parallel resistors: I = pd across the parallel resistors

resistance of each resistor

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Single cell questionTotal resistance of the circuit= 8 Ω in series with 12 Ω in parallel with 6 Ω= 8 + 5.333 = 13.333 Ω

Total current drawn from the battery = V / RT

= 9V / 13.333 Ω= 0.675 Apd across 8 Ω resistor = V8 = I R8

= 0.675 A x 8 Ω = 5.40 Vtherefore pd across 6 Ω (and 12 Ω) resistor, V6 = 9 – 5.4pd across 6 Ω resistor = 3.6 V Current through 6 Ω resistor = I6 = V6 / R6 = 3.6 V / 6 Ω current through 6 Ω resistor = 0.600 A

8 Ω

9 V

12 Ω

Calculate the potential difference across and the current through the 6 ohm resistor in the circuit below.

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