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AL Current Electricity P.66

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Current = rate of flow of charge through cross-sectional area I = d Q / d t I = 1 C / 1 s = 1 C s -1 = 1 A One coulomb is the quantity of electric charge for a steady current of 1 ampere flowing through a given point for 1 second.

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P.66 Q = I t = area under the curve Q = area = (1/2)(5)(5)+(5)(7-5)+(-4)(10-7) Q = area = 10.5 (C)

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P.66 Conductors => mobile electrons, move freely in electrons sea Insulators => no mobile electrons, strictly hold by atoms

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P.66

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P.67 The steady average velocity of the free electrons in the drifting direction or, effectively, in the direction opposite to that of the electric field is known as the drift velocity of the free electrons in the metal. Drift velocity

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P.67 Drift velocity Current = Rate of flow of charge I = nAvQ

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P.67 Mechanism of electrical conduction in metals Notes: 1. For copper, since the value of the number of free electrons per unit volume (n) is large (10 29 ), the electron drift velocity (v) is rather small. It is less than 1 mm per second. 2. At higher temperatures, the amplitude of vibration of the ions increases and the free electrons collide more frequently with the ions. The drift velocity of the free electrons decreases. Thus, the current decreases.

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P.68 Distinction between drift velocity and speed of electrical signal 1. Average random speed: - averaging random speeds in a specific direction - order of magnitude of 10 6 m s -1 2. Drift velocity: - electrons travels from one end to the other end of a wire - electric field is set up inside - much smaller order of magnitude of 10 -4 m s -1 3. Electrical signal speed: - speed of electric field - = speed of EM waves (3 x 10 8 m s -1 )

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P.68 Average random speed Drift velocity Electrical signal speed CarrierElectrons Electric field ConditionNo current flow Current flows Order of magnitud e / m s -1 10 6 10 -4 10 8

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P.68

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P.69 Mass of 1 mole = 64 g = 0.064 kg No. of mole in 1 m 3 = 9000 / 0.064 = 140625 No. of atom in 1 m 3 = 140625 (6x10 23 ) = 8.4375 x 10 28 No. of free e - in 1 m 3 = n = 8.4375 x 10 28

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P.69 Conduction electrons collide the lattice ions only due to higher chance.

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P.69 Electromotive force (e.m.f.) Electric current from P (higher potential) to Q (lower potential) - until P and Q at same potential

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P.69 Electromotive force (e.m.f.) The electromotive force (e.m.f.) of a source is defined as the electrical potential energy supplied to each coulomb of charge by the source to drive the charge around a complete circuit. Unit: volt (1 V = 1 J C -1 )

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P.69 Potential difference (p.d.) The potential difference (p.d.) between two points in a circuit is the energy converted from electrical potential energy to other forms when one coulomb of charge passing between the points outside the power source. Unit: J C -1 or volt (V)

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P.69 The p.d. between two points in a circuit can also be defined as the ratio of the power dissipated between the two points to the current that flows through the two points.

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P.69 Potential difference (p.d.) 1. Definition of volt (1 V) - 1 J of electrical energy is converted to other forms of energy between two points when 1 C of charge flows - power dissipated between two points is 1 W when current flowing is 1 A

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P.69 Potential difference (p.d.) 2. Measurement of p.d.: - voltmeter is connected across two points

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P.70 Combination of cells 1. In series = 1 + 2 + 3 2. In parallel the equivalent e.m.f. of the system is equal any one of the e.m.f. of cells. =Max( 1, 2, 3 )

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P.70 Electrical resistance 1. Definition of resistance - unit: ohm ( ) The ohm (1 Ω) is the resistance of a conductor when a current of one ampere (1 A) flows through the conductor, the potential difference across it is one volt (1 V).

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P.70 Current – voltage relation

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P.70 Ohm’s Law Ohm’s Law states that the steady current through a metallic conductor is directly proportional to the potential difference across it, provided that the temperature and other physical conditions remain constant. Ohmic conductor - conductor obeys Ohm’s Law (I p.d.)

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P.70 Non-ohmic conductor - I-V characteristics are not straight lines Notes: 1. The expression V/I = R is not a representation of Ohm’s Law but a representation of the resistance R. A non-ohmic conductor also has a resistance which can be found using the equation R = V/I but its value is not a constant. 2. Ohm’s Law can be represented by the expression V/I = constant only when the temperature is constant. This implies that at constant temperature, the resistance of an ohmic conductor is independent of the current I or the potential difference V. 3. Therefore, Ohm’s Law is only a special case of resistance behaviour.

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P.70

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P.72 Combination of resistors Resistors in series In series For the resistors connected in series, the equivalent resistance (R) is:

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P.72 Combination of resistors Resistors in parallel In parallel 1. Equivalent resistance For the resistors connected in parallel, the equivalent resistance (R) is:

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P.72 Resistors in parallel Notes: For resistors joined in parallel, the equivalent resistance is always less than the resistance of any one of the resistors. 2. Currents in resistors

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P.72 3. High resistance of voltmeter R 2 >> R 1, V’ V

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P.72 Combination of resistors http://lectureonline.cl.msu.edu/%7Emmp/kap20/RR506a.htm

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P.73 Factors affecting resistance of a conductor Physical dimension Material Effect of temperature on resistance

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P.73 Physical dimension

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P.73 Material Different materials have different conducting properties - resistivity = unit: m

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P.73 MaterialResistivity Silver1.62x10 -8 Copper1.69x10 -8 Tungsten5.25x10 -8 Platinum1.06x10 -8 Silicon-pure2.5x10 3 n-type8.7x10 -4 p-type2.8x10 -3

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P.73 Effect of temperature on resistance 1. Effect on conductors - In solids, atoms vibrate about their equilibrium positions - temperature increases, amplitude of vibration larger - chance of collision with free electrons higher - resistance higher - current lower

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P.73 2. Temperature coefficient of resistance R - resistance at temperature = R 0 (1 + ) Notes: One of the features of temperature coefficient of resistance is that for many metals its value is close to 0.003 66 o C –1, which is the reciprocal of 273 o C.

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P.73 r = internal resistance of the cell Internal resistance of a cell and terminal voltage

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P.73 Internal resistance of a cell and terminal voltage Determining internal resistance of a cell terminal voltage - voltage across the terminals of a cell ε = V + Ir V= (-r) I + ε Slope = -r

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P.74 73 Ω Resistance Box

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P.74 Potential Divider R2R2 R1R1 A C VoVo V out

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P.74 Effect of external load (R L ) on output p.d. If R L >> R 1, R E = R 1 V’ = voltage across R 1 1. Rheostat – provide continuously variable p.d. from 0 to full voltage of supply V

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P.74 2. Thermistor – resistance decreases with increase in temperature

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P.75 (1) F T T U = QVIn parallel connection, p.d. are the same Q are the same, so U are the same. (2) (3)

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P.75 Switch S is open Consider the division of voltage G K Switch S is closed Consider the division of voltage

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P.75 When cross-sectional area A is increasing, then the drift velocity is decreasing

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P.75 v

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R 1 = V / I = 1 ΩR 2 = V / I = 2 / 5 = 0.4 Ω ΔR = R 2 - R 1 = 0.4 – 1 = - 0.6 Ω

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P.76 Maximum power

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P.76 Efficiency For maximum power output R = r, = 50%

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P.77 For running motor, back emf εwill be induced T T T V – ε= I R I V – Iε= I 2 R = power dissipated as heat

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P.78 R X = 6 2 /12 = 3ΩR Y = 6 2 /3 = 12Ω In order to operate at rated value, both p.d. should be 6V. Besides, the voltage supply is 12V, so each of them should be shared 6V in series connection. X X C. If they are connected in series, current will be the same, but with different resistances will have different voltage. X In parallel connection, the equivalent resistance will be smaller than any one to the resistor in connection. It is possible to reduce the equivalent resistance to 3Ω

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P.78 The bulbs are non-ohmic Total p.d. of bulbs are 200 V The current flowing through bulbs are 0.24 A For X, I = 0.24 A and V = 50 V P X = (0.24)(50) = 12 (W) For Y, I = 0.24 A and V = 150 V P X = (0.24)(150) = 36 (W)

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P.78 L x = 3 L y m x = m y A x L x = A y L y 3 A x = A y P = I 2 R

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P.78 At y-intercept, R=0

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P.78 (1) The glass wall of bulb ONLY absorb heat released by the filament, brightness is NOT affected. F F T (2) Power taken by filament should be dominated by the very large resistance of filament, thus less energy is lost in other part of the circuit. (3) The filament emits visible light only when it is very hot. Most of them are infra-red radiation.

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P.79 In parallel connection, p.d. is the same. R 2 = 3 R 3 Equivalent resistance in parallel connection P 1 = P 2

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P.79 When switch S is closed, the equivalent resistance across R 1 and R 2 will be reduced. More p.d. across R 3. Potential at point Q will be more negative. G Potential at point P will be less positive.

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P.79 (1) If r > R, then greater p.d. across r than R. T T T The terminal p.d. (ε – I r) = I R will be small. Power loss I 2 r of the battery will be greater. (2) If R > r, V R will be greater. Power loss I 2 R of the resistor will be greater. (3) If R = r, Power supplied by battery will be maximum.

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P.79

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P.80 For maximum power dissipated, Total external R = internal resistance r R + 6 = r It is impossible to have solution R= 0 can have larger power consumption

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P.80

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P.81

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First Law (Current Law) Kirchhoff’s Laws – find currents flowing in different parts of a network The algebraic sum of the currents at a junction of a circuit is zero. Therefore, the current arriving a junction equals the current leaving the junction. i.e. ΣI = 0 Current +ve – flows into a point -ve – flows out from the point

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P.81 Second Law (Voltage Law) (Optional) Round a closed loop, the algebraic sum of the e.m.f.s is equal to the algebraic sum of the products of the current and resistance. i.e. ΣE = Σ (IR)

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P.81 V x is equal to e.m.f. 2V X No voltmeter reading

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P.82 V A – V C = (0.1)(1) = 0.1 0.1A A B C V A – V B = (0.2)(5) = 1 If V C > V B, current flows from C to B. I1I1 V C - V B = I 1 (3) (V A - 0.1) – (V A – 1) = I 1 (3) I 1 = 0.3 I2I2 By Kirchhoff’s 1st law, I 2 = I 1 – (0.1) I 2 = 0.2 (A)

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P.82 By Kirchhoff’s Law A B C D

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P.82 C D R R R 1k S C A R R S

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P.83 Draw most of voltage across R Draw most of current through S

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P.83 Conversion to ammeter Limit f.s.d. current Connect a small resistor (shunt) in parallel to draw majority of current In ideal case, internal resistance of ammeter is zero.

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P.84 Conversion to voltmeter Limit f.s.d. voltage Connect a large resistor (multiplier M) in series to share majority of p.d. In ideal case, internal resistance of voltmeter is infinite.

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P.84 For 10A mode 9.9A (9.9) R 1 = (0.1) (10+R 2 ) 0.1A For 1A mode (0.9) (R 1 +R 2 ) = (0.1) (10) 0.9A 0.1A Solve it (0.9) (R 1 +R 2 ) = (0.1) (10) (0.9) (R 1 +1) = 1 R 1 = 1/9

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P.84 For this conversion, a multiplier (high resistance) should be connected in series with the galvanometer to share most of the input voltage. 5 = (100μ)(1k + R M ) 5 = (100μ)(R total ) R total = 50 kΩ

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P.85 True reading : If R >> R A, error in voltmeter reading is relatively smaller Ammeter Wrong reading : Voltmeter Measured R : Suitable for LARGER R True reading : If R << R V, error in ammeter reading is relatively smaller Voltmeter Wrong reading : Ammeter Measured R : Suitable for SMALLER R

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P.86 As Voltmeter

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P.86 As Ammeter

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P.86 As Ohmmeter

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P.86 S is open By division of voltage A C R SR V ε A C R SR V ε S is closed R is shorted V’ = ε

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P.86 (1) Some current flowing through the voltmeter, the reading of ammeter is greater than true value. T T F (2) Voltmeter reading is accurate, but the ammeter reading is greater than true value. So, the ratio of V /I is smaller than the true value. (3) If the resistance of R is large and compatible to the internal resistance of voltmeter. The current flowing through the voltmeter is not negligible. The measured value of R is wrong.

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By division of voltage, P.87 60 Ω resistors are neglected. V 10 Ω resistors are neglected. V

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P.87 V By division of voltage, 1 Ω resistors are neglected. A By division of voltage,

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P.87

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P.88

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P.87

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P.88 Wheatstone Bridge If P, Q, R are given, Put S into the circuit and the switch is closed If the reading of galvanometer is ZERO, It means that V B = V D By division of voltage, Voltage ratio in ABC = Voltage ratio in ADC

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P.89

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At steady state, the capacitor is fully charged V C = V B – V A No current flowing through capacitor By division of voltage, Q = C(V B – V A ) Q = (20x10 -6 )(4 – 2) Q = 40x10 -6 Q A = -40x10 -6 Lower in potential

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P.89 Metre Bridge Metre bridge – simple form of Wheatstone bridge

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P.89 Metre Bridge If the reading of galvanometer is ZERO,

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P.90 When the bridge is balanced, Notes: When using the metre bridge, certain precautions need to be taken. 1. The value of the standard resistance S need to be chosen such that the balance point B is about in the middle third of the slide wire, i.e. roughly between the 30 cm mark and the 70 cm mark. This is done by trial and error. When the balance point falls in the middle third of the wire, the percentage errors in the measurement of l 1 and l 2 are of the same order.

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P.90 Notes: 2. The slider should not be slided on the slide wire so that its uniformity is maintained. 3. The experiment must be repeated with the unknown resistance R in the right-hand gap and the standard resistance S in the left-hand gap. This is to eliminate end errors. A source of end error is the point where the slide wire is soldered to the copper strips. The different amounts of solder used may result in the points of contact having different electrical resistances. Metre bridge – not suitable for low resistance

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P.90 Potentiometer If the reading of galvanometer is ZERO, Metre Bridge V 1 = V XY = V AC (1) Protect galvanometer when large current flowing through it. (2) Close the switch to have more accurate reading

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P.91 Measurement of p.d. If the slider moves towards point B, Potential at point C < Potential at point Y Current flows from point Y to point C

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P.92 Calibrating a voltmeter Read the reading of point C Voltmeter reading should be the same, otherwise Adjust the value of rheostat until the reading of galvanometer is zero

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P.93 Calibrating an ammeter Read the reading of point C Ammeter reading should be the same, otherwise Adjust the value of rheostat until the reading of galvanometer is zero

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P.93 Measurement of internal resistance of a cell Read the reading of point C when the switch is open. V = e.m.f. of cell Internal resistance r is useless because No current flowing through cell and r Read the reading of point C when the switch is closed. Plot a graph of (1/l) against (1/R)

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P.93 Measurement of internal resistance of a cell intercept on 1/R- axis

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P.94

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Electric Currents and Resistance

Electric Currents and Resistance

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