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**BHS Physical Science K Warne**

Electricity BHS Physical Science K Warne

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Grade 9 Revision Revision Presentation

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**Voltage across …………..= voltage in ……………**

Electrical Circuits The Ammeter measures the ……………flowing in the circuit. (…….. A) The Voltmeter Measures potential difference or …………….. in volts. (V) The Resistance of the Resistor is given in …………… (Ω). V1 = V2 V1 Voltmeter A Ammeter Resistor V2 Voltage across …………..= voltage in ……………

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**Electric Current in a Conductor**

- + Conventional current - __________________ Direct current - moves in _______________. Alternating current - ___________________ continuously Maintaining a current Conductor - ___________ _____________________ _____________________ - SOURCE e- Conventional current - + e- e- e- e- e- + + + + + + + + + + e- e- e- e- e- e- e- + + + + + + + + + + + e- e- - e- e- + e- + + + + + + + + + e- e- e- e- e- + + e- + + + e- + + + + + < electrons Conventional current is the movement of from + to - in a conductor. “_____________” >

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**Electric Current & Voltage**

- + The VOLTAGE is how much ENERGY they have. e- Conventional current - + Current is the number of soldiers moving past a point.

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**Potential Difference _ + W or E (J) ………….(J) Q (C ) V = ………….(C )**

The DIFFERENCE in POTENTIAL (energy) per unit ………………. of the current flowing between two points in the circuit. Measured by a ………………... W or E (J) Q (C ) ………….(J) ………….(C ) Voltmeter V = Volts = V 1.50 + _ Resistor The Voltmeter: is connected in ……………. to another component in the circuit (the current does NOT flow through it) has a …………. resistance Is connected positive to positive - negative to negative. V R

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**POTENTIAL DIFFERENCE A ? A 2V ?? Energy lost or work done?? Example 1:**

Calculate the potential difference between two points if 20 J of work are required to move a charge of 2 C. Example 2: Calculate the work done in moving a charge of 5 C through a potential difference of 2 V. A ? 2 coulombs of charge Resistance R 20J Energy lost A 2V 5 coulombs of charge Resistance R ?? Energy lost or work done??

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**EMF - Electro Motive Force**

Vex A Resistance R VCell Emf 400V Emf is the …………… amount of …………… that the cell can produce (per unit charge). Measured when the current in the circuit is ……….. Open Circuit!! Vcell = EMF I = 0A Vcrt.= ….. v Emf = …..cell

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**External Potential Difference**

………… Circuit!! 50V Energy …………….. by battery is lost by resistance in the circuit. The EMF of the cell is equal to the sum of the ………………..cell voltage and the ………… voltage. This continues until the cells have no more energy. Vex A Resistance R Vcell I = IA (EMF = 400V) 350V 50V Vex= Vcell 400V 350V Emf = V….. + V……

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**Measuring Current Ammeter _ + A 1 I = Q/t 1A = …..C/……s A**

An Ammeter measures the ……………… flowing through the circuit. Ammeter _ A 1 + The current is the number of charges passing a point in one second. I = Q/t 1A = …..C/……s The Ammeter: is connected in ……….in the circuit (the current flows through it) has a …………resistance Is connected positive to positive - negative to negative. A

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**Calculating Current Ammeter _ = + A 0.53 I = Q/t**

Calculate the current flowing through the circuit. Ammeter A _ 0.53 + If 160 C of charge flow through the ammeter in 3s what current is flowing? I = Q/t =

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**Parallel Circuits 1 1 1 Rt R1 R2 = + Rt … … 1 1 1 1 Rt 2 2 2 = +**

Adding resistors in parallel…decreases the total resistance. Rt R1 R2 Rt … … Rt Rt = …./…. = ……. Ω Total R = …. Ω R1 2 Ω = + R1 2 Ω = + R2 2 Ω Total R = … Ω = R1 2 Ω R2 2 Ω R2 2 Ω Total R = …… Ω

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**Parallel Circuits VT = ……………. AT= ……………….. A A1 A2**

The voltage is EQUAL over the resistances . VT = ……………. The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow. 2 V VT V1 ….. V R1 V2 ….. V R2 4 A A R1 AT= ……………….. A1 …. A A2 ….. A R2

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**Parallel Circuits 1 1 1 Rt R1 R2 = + VT = V1 = V2**

Adding resistors in parallel…decreases the total resistance. Rt R1 R2 The voltage is EQUAL over the resistances . VT = V1 = V2 The current flowing is divided between the resistances and would increase as more resistances are added- more routes for the current to flow. VT A = + V1 2 Ω R1 A1 V2 A2 R2 2 Ω The current will divide in such a way that the potential lost by both all branches of current will be the same. AT= A1 + A2

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**The potential differences will be proportional to the resistances.**

Series Circuits 12v Adding resistors in series…increases the total resistance - because all the current flows through all the resistors. Rt = R1 + R2 The total potential difference (voltage) is the sum of the potential differences of the resistors – the total potential loss must equal the all the potential lost along the way. Vt = V1 + V2 The potential differences will be proportional to the resistances. The current flowing is the same all over the circuit and would decrease as more resistances are added - A = A2 = A3 VT V1 v2 A A2 A3 1Ω 3Ω

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Worked Example 4.1 A 4 Ω 2 Ω 8 Ω 12 Ω 6 Ω 12v Calculate I

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**Current & Resistance _ +**

CURRENT: An electrical current is a movement of ……………. through a conducting material from positive to negative. (?!) _ + RESISTANCE Electrical charge experiences ……………………as it moves through a conductor. The resistance is due to ……………….. with particles in the metal atoms and ions. The moving charges lose ……………………in the collisions which …………….up the conductor.

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**Ohm’s Law - Practical AIM: A V**

Investigate the relationship between the …………………….. across a resistor and the ………………..flowing through it. Determine the …………………..of a resistor. rheostat METHOD: Set up the circuit as shown. Using the rheostat vary the current in the circuit, obtain a range of readings for the potential difference across R for different currents. RESULTS>> A V Resistance R

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**Ohm’s Law - Results Analysis - Graph Results**

I (A) V (V) Draw a graph of your results.

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Ohm’s Law - Analysis The ratio V/I produces a constant value - for any resistor This is the Resistance of the resistor. The Unit of measurement for resistance is the Ohm - symbol (Ω) The SLOPE of the graph gives the RESISTANCE. Rise run Slope = DY/ DX =

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**Current, voltage & resistance**

+ - A V The relationship between the …………… through a resistor, the …………... drop across the resistor and the resistance of the resistor is expressed by the following equation: R V Calculate the voltage drop across a 2 resistor when a current of 1.5 A is flowing.. Rx .. We define the unit of resistance; one …….() is one volt per ampere. R = V/I 1 = …V/…A

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**Worked Example 4.1 A Calculate I = V/R = 12/6 = 2 A 12v 6 Ω 4 Ω 12 Ω**

8 Ω 12 Ω 6 Ω 12v Calculate I = V/R = 12/6 = 2 A

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**Ohm’s Law - Factors affecting Resistance Material + Length Temperature**

V

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**Effects of Current _ _ + +**

Electric current generates heat in a conductor. + _ A small current (0.1A) would have only a few charges flowing. + _ A large current (15A) would have a large number of charges flowing and generate far more heat. As a conductor heats up the RESISTANCE INCREASES.

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**Effects of electric current**

An electric current that flows in a conductor has a number of effects: HEATING The friction caused by the current causes the conductor to heat up. The greater the current the more heat is generated. MAGNETIC EFFECT - A magnetic field is generated around any conductor when an electric current flows through it.

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Ohm’s Law PSSA Requirement Unit 9 Honors Physics.

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