AL Current Electricity P.66

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

P Ω Resistance Box

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

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)

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

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

P.75 v

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

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

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

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)

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

P.78 At y-intercept, R=0

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.

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

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.

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.

P.79

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

P.80

P.81

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

P.82 V A – V C = (0.1)(1) = A 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)

P.82 By Kirchhoff’s Law A B C D

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

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

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

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

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

P.86 As Voltmeter

P.86 As Ammeter

P.86 As Ohmmeter

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

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.

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

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

P.87

P.88

P.87

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

P.89

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

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

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

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

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

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

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)

P.94