Electric current g11

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Electric Current

Keith WarneGrade 11 & 12

http://www.warnescience.net/



Prior Knowledge Summary

Ix = (ItR//)/Rx

Resistors in series potential dividers

Parallel resistors

Current dividers




Electric Current in a Conductor

• Conventional current -

positive to negative

• Direct current - moves in

one direction.

• Alternating current -

changes direction

continuously

• Maintaining a current

– Conductor

– Potential difference

– Replacement of charges

+ -

+ -

<------- electrons

“Positive spaces” ------>Conventional current is the movement of from

+ to - in a conductor.

+ +

+

+

+

+

+ +

+ +

+

+

+

+

+ +

+ +

+

+

+

+

+ +

+ +

+

+

+

+

+ +

+ +

+

+

+

+

+ +

e-

e-

e-e-

e-

e-

e-e-

e-

e-

e-e-

e-

e-

e-e-

e-e-

e-e-+ -e-

e-

e-

<------- electrons

e-




Ohm’s Law - PracticalAIM:

– Investigate the relationship between the potential difference across a

resistor and the current flowing through it.

– Determine the resistance of a resistor.

V

A

METHOD:

1. Set up the circuit as shown.

2. Using the rheostat to vary the

current in the circuit, obtain a

range of readings for the

potential difference across R for

different currents.

Resistance R

rheostat




Ohm’s Law - Results

The graph is a straight line showing that the current (I) is

directly proportional to the voltage (V).

I (A) V (V)

0.80 3.40

0.81 3.50

0.85 3.70

0.90 3.90

0.95 4.20

Current vs Voltage

3.00

3.20

3.40

3.60

3.80

4.00

4.20

4.40

0.75 0.80 0.85 0.90 0.95 1.00Current (A)

Vo

ltag

e (V

)

ResultsAnalysis




Worked Example

A

4 Ω

2 Ω

8 Ω

12 Ω

6 Ω

12v

Calculate I = V/R = 12/6 = 2 A




Energy & Power

Work = ……… = ………= …….. = Energy

Power = the rate at which work is done

= ………………..

= ……….

= ……..

=………

From Joules experiment


Joules Experiment.notebook



Power Investigations

Design plan and conduct a simple investigations into the power output of

bulbs (or resistors) when connected in series and parallel.

The questions you need to answer are:

1) How is the power output of a bulb (or resistor) affected when other

similar bulbs are connected to it;

a) In series

b) In parallel

2) How is the potential difference across the power supply (battery)

affected when the resistance in the circuit is changed.

CRITERIA

You may use circuit boards and croc clips simulations to gather your data.

(at least one investigation must be done using a circuit board)

You must take measurements of current and voltage and prove your

conclusions with calculations and observations of bulb brightness.

Circuit diagrams must be drawn and method explained as well as any

relevant graphs drawn.

Hand in: Plan, Aim, Hyp, Method, (Incl Diagrams), Results (Table),

Analysis (Calculations), Conclusions




Series Power

• Increasing the resistance in the circuit decreases the current.

• Lower current flowing through each bulb results in a lower

voltage drop and therefore less power is dissipated.

• . (No internal resistance) The voltage drop across the whole

circuit remains constant




A non ohmic resistor (bulb) under varying voltage

POWER (W)

Volts (V)

CURRENT (A)

RESISTANCE (W)

POWER (W)

BULBS R = V/I P=V.I

6 12 0.5 24.0 6.0

6 10 0.453 22.1 4.5

6 8 0.402 19.9 3.2

6 6 0.343 17.5 2.1

6 4 0.274 14.6 1.1

Non Ohmic Conductor/Resistance




P10 = VI

= (3.41)(0.418)

= 1.4 W

P6 = VI

= (8.59)(0.418)

= 3.6 W

Series Bulb BrightnessThe higher resistance of the 6 W bulb will limit the current available so

the 6 W bulb will deliver more power than the 10 W bulb when in

series.

Higher resistance gives more power series! Since current equal in

series the one with the highest voltage (highest R) has most power.

R10w = V/I = 3.41/0.418 = 8.2W

R6w = V/I = 8.59/0.418 = 20.6 W




Electricity Questions - SG

4.1 The battery has negligible

internal resistance.

Calculate:

4.1.1 The total resistance of the

circuit. (4)

4.1.2 The current in the 2.4 W

resistor. (3)

4.1.3 the potential difference

across the 2.4 W resistor.

(3)

4.1.4 the potential difference

across the parallel combination.

(2)

A

12V

S

2.4W

XX

X2W

8W




Vex

A

Resistance R

VemfEmf

400V

I = 0A

Vex= 0 v

Emf = …………..

• Emf is the

…………………

amount of energy

that the cell can

produce (per unit

charge).

• Measured when the

current in the circuit

is …………..

EMF - Electro Motive Force

Open Circuit!!




Infinite R

No current

voltage = EMF

Large R

Small current

voltage < EMF

Small R

BIG current

voltage << EMF

Lost volts

increase!

Lost volts

small

amount!

DECREASING RESISTANCE in the circuit DECREASES

VOLTAGE drop in the circuit!

Open

circuit




Exam QuestionsQUESTION 1 One word or phrase answers

1.1The law which relates the current in a resistor, maintained at constant

temperature, to the potential difference across its ends. (1)

1.2The unit of measure equivalent to one volt per ampere. (1)

Reading on V1 Reading on V2

A 12 V 0 V

B 12 V 12 V

C 0 V 0 V

D 0 V 12 V

QUESTION 2 Multiple Choice

2.1 A variable resistor, an ammeter, a battery of emf 12 V and voltmeters V1

and V2 are connected as shown in the diagram below.

When the switch is open, the readings on voltmeters V1 and V2 respectively are …




Electricity QuestionsThe battery in the circuit diagram below has an EMF of 12 V and an unknown internal

resistance r. Voltmeter V1 is connected across the battery and voltmeter V2 is connected

across the switch S. The resistance of the connecting wires and the ammeter is negligible.

1 Write down the respective readings on

voltmeters V1 and V2 when switch S is

open. (2)

Switch S is now closed.

The reading on voltmeter V1 changes to 9 V.

2. What will the new reading on V2 be?(1)

3. Calculate the total external resistance of

the circuit. (4)

4. Calculate the internal resistance, r, of the

battery. (5)ANSWERS >>




Hi -

This is a SAMPLE presentation only.

My FULL presentations, which contain a lot more more slides and other resources, are freely

available on my resource sharing website:

www.warnescience.net(click on link or logo)

Have a look and enjoy!

WarneScience





Electric current g11

Education

series power

power series

resistance w power w

direct current

w resistor

measurements of current

current vs voltage

w series bulb brightness