AL Current Electricity P.66 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.

AL Current Electricity

P.66

P.66Current = 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.

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)

P.66Conductors => mobile electrons, move freely in electrons sea

Insulators => no mobile electrons, strictly hold by atoms

P.66

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

P.67Drift velocity

Current = Rate of flow of charge t

QAvtnI

Time

Charge

I = nAvQ

P.67Mechanism of electrical conduction in metals

Notes:

1. For copper, since the value of the number of free electrons per unit volume (n) is large (1029), 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.

P.68Distinction between drift velocity and speed of

electrical signal1. Average random speed:- averaging random speeds in a specific direction- order of magnitude of 106 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 108 m s-1)

P.68

Average random speed

Drift velocity

Electrical signal speed

Carrier Electrons Electrons Electric field

Condition

No current

flow

Current flows

Current flows

Order of magnitude / m s-

1

106 10-4 108

P.68

nqvA

I

nQvvQnA

I422

nQvvQnA

I1010

nQvvQnA

I33

P.69

Mass of 1 mole = 64 g = 0.064 kg

No. of mole in 1 m3 = 9000 / 0.064 = 140625

No. of atom in 1 m3 = 140625 (6x1023) = 8.4375 x 1028

No. of free e- in 1 m3 = n = 8.4375 x 1028

P.69

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

P.69

Electromotive force (e.m.f.)

Electric current from P (higher potential) to Q (lower potential) - until P and Q at same potential

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.

charge

suppliedenergy source a of e.m.f.

Unit: volt (1 V = 1 J C-1)

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)

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.

I

PV

IVt

IVtP

IVtQV

Time

dissipatedenergy Electricalpower Electrical

dissipatedenergy Electrical

Charge

dissipatedenergy Electricalp.d.

P.69

Potential difference (p.d.)

Q

EV

QVE

p.d. Charge Energy

I

PV

IVP

p.d.Current Power

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

P.69

Potential difference (p.d.)

2. Measurement of p.d.:- voltmeter is connected across two points

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 )

P.70

Electrical resistance1. Definition of resistance- unit: ohm ()

I

VR )( Resistance

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).

P.70

Current – voltage relation

P.70

Ohm’s LawOhm’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.)

P.70Non-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.

P.70

P.72 Combination of resistorsResistors in seriesIn series

321

321321 ,

RRRI

V

IRIRIRVVVVV

For the resistors connected in series, the equivalent resistance (R) is:

...221 RRRR

P.72 Combination of resistorsResistors in parallel

In parallel1. Equivalent resistance

321

33

22

11321

111

,,,

RRRV

R

V

R

VI

R

VI

R

VIIIII

For the resistors connected in parallel, the equivalent resistance (R) is:

P.72

Resistors in parallelNotes: For resistors joined in parallel, the equivalent resistance is always less than the resistance of any one of the resistors.

2. Currents in resistors

currentMain sresistance of Sum

resistanceother The

sresistance of Sum

sresistance ofProduct

21

21

21

2111

21

21

IRR

RI

RR

RRIIRRI

RR

RRR

P.72

3. High resistance of voltmeter

VRR

R

RR

RRIIRVV

RR

RRIV

RR

RRR

21

1

21

211

21

21

21

21

'

',

R2 >> R1, V’ V

P.72

Combination of resistors

http://lectureonline.cl.msu.edu/%7Emmp/kap20/RR506a.htm

P.73

Factors affecting resistance of a conductorFactors affecting resistance of a conductor

Physical dimensionPhysical dimension

MaterialMaterial

Effect of temperature on Effect of temperature on resistanceresistance

P.73

Physical dimension

AR

)( Resistance

P.73

Material

Different materials have different conducting properties

AR

- resistivity =

unit: m RA

P.73

Material ResistivitySilver 1.62x10-8Copper 1.69x10-8

Tungsten 5.25x10-8Platinum 1.06x10-8

Silicon-pure 2.5x103n-type 8.7x10-4p-type 2.8x10-3

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

P.73

2. Temperature coefficient of resistance

0

0

resistance oft coefficien eTemperatur

R

RR

R - resistance at temperature

= R0 (1 + )Notes: One of the features of temperature coefficient of resistance is that for many metals its value is close to 0.003 66oC–1, which is the reciprocal of 273oC.

P.73

r

VI

r = internal resistance of the cell

Internal resistance of a cell and terminal voltage

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 + IrV= (-r) I + ε

Slope = -r

P.74

73 Ω

Resistance Box

P.74

Potential Divider

R2

R1

A

C

21

0

RR

VI

Vo

Vout

021

111 V

RR

RIRVVout

P.74

Effect of external load (RL) on output p.d.

1L1

L1E

L1E

111

RRR

RRR

RRR

If RL >> R1, RE = R1

V’ = voltage across R1

1. Rheostat – provide continuously variable p.d. from 0 to full voltage of supply V

P.74

2. Thermistor – resistance decreases with increase in temperature

VRR

RV

21

1' VRR

RV

21

2'

P.75

321

1111

RRRR

(1)

F

T

T

U = QV In parallel connection, p.d. are the same

Q are the same, so U are the same.

(2)

2

11

RR RR 2

(3)

32

111

RRRnew

R

1 RRnew

P.75

)(26)42(

2VVVV PGGP

Switch S is open

Consider the division of voltageGK

)(2 VVP

Switch S is closed

Consider the division of voltage

)(36)

)44()4)(4(

2(

2VVVV PGGP

)(3 VVP

P.75

nAq

Iv

2

1

Anq

I

dA

dv

When cross-sectional area A is increasing,

then the drift velocity is decreasing

P.75v

nAq

Iv

A

lR

IRV

lvnqnqvlA

lIIRV )()(

P.75

R1 = V / I = 1 Ω R2 = V / I = 2 / 5 = 0.4 Ω

ΔR = R2 - R1 = 0.4 – 1 = - 0.6 Ω

P.76

Maximum power

rR

rR

rRRErRE

dR

dP

RrR

EIVP

rR

EI

)(

20

power) maximum(For

,

4

222

2

P.76

Efficiency

rRI

RI

2

2

inputPower

outputPower

%100

rR

R

For maximum power outputR = r, = 50%

P.77

For running motor, back emf εwill be induced

T

T

T

V – ε= I R

I V – Iε= I2 R = power dissipated as heat

P.78

RX = 62/12 = 3Ω RY = 62/3 = 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.

XX

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

PX = (0.24)(50) = 12 (W)

For Y, I = 0.24 A and V = 150 V

PX = (0.24)(150) = 36 (W)

P.78

Lx = 3 Ly

A

lR

mx = my

Ax Lx = Ay Ly

3 Ax = Ay

x

xx A

LR

y

yy A

LR

1

9)3)(3(

xy

yx

y

x

AL

AL

R

R

P = I2 R

4

9

)2(

)1(2

2

y

x

y

x

R

R

P

P

P.78

At y-intercept, R=0)( ArrRI

)5.0(5.1 rIIR

5.1

)5.0(

5.1

11 r

RI

5.1

)5.0(88.0

r

)(82.0 r

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

VP

2

3

1

3

2

2

3 P

P

R

R R2 = 3 R3

Equivalent resistance in parallel connection

332

3223 4

3R

RR

RRR

12

1 RIP

)3()3

()( 32

33

32

2

32

32 RI

RR

RRI

RR

RP

32

2 16

3RIP P1 = P2

12

32

2 16

3RIRIP

31 16

3RR

P.79

When switch S is closed, the equivalent resistance across R1 and R2 will be reduced.

More p.d. across R3.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 I2 r of the battery will be greater.

(2) If R > r, VR will be greater. Power loss I2 R of the resistor will be greater.

(3) If R = r, Power supplied by battery will be maximum.

P.79

)(

2

rR

VP

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.80

P.80

P.80

P.80

P.81

P.81

P.81

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

P.81Second 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)

P.81

Vx is equal to e.m.f. 2V

2)6(105

5

kk

kVX

X

No voltmeter reading

P.82

VA – VC = (0.1)(1) = 0.1

0.1A

A

B

C

VA – VB = (0.2)(5) = 1

If VC > VB , current flows from C to B.

I1

VC - VB = I1 (3)

(VA - 0.1) – (VA – 1) = I1(3)

I1 = 0.3

I2

By Kirchhoff’s 1st law,

I2 = I1 – (0.1)

I2 = 0.2 (A)

P.82

By Kirchhoff’s Law

rI AC

4

A B

C

D

549 CDV

rICD

5

CDBCAC III

rI

r BC

54

rIBC

1

1 rIV BCBC

651 CBCB VVV

369 BAAB VVV

P.82

SkRRRReq

1

1111

C

D

RR

R

1k

S

Rk

2

5.2

1 kR 5

C

A

R

R

1k S

11

1

1'

SkRR

RRReq

kkk

kR

Req 5.312

51

2'

P.83

Draw most of voltage across R Draw most of current through S

P.83Conversion 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.

P.84Conversion 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.

P.84

For 10A mode

9.9A

(9.9) R1 = (0.1) (10+R2)

0.1A

0.1A

For 1A mode

(0.9) (R1 +R2) = (0.1) (10)

0.9A 0.9A

0.1A

Solve it

(0.9) (R1 +R2) = (0.1) (10)

19.9

)10(1.09.09.0 2

2

R

R

111.019.9 22 RR

12 R

(0.9) (R1 +1) = 1

R1 = 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 + RM)

5 = (100μ)(Rtotal)

Rtotal= 50 kΩ

P.85 True reading :

If R >> RA, error in voltmeter reading is relatively smaller

RA

R

A C E

B D F

A

V

RV

Figure 1

RA

R

A C

EB

D

F

A

V

RV

Figure 2

Ammeter

Wrong reading : Voltmeter

Measured R :t

m

m

mm I

V

I

VR t

t

t RI

V

Suitable for LARGER R

True reading :

If R << RV, error in ammeter reading is relatively smaller

Voltmeter

Wrong reading : Ammeter

Measured R :m

t

m

mm I

V

I

VR t

t

t RI

V

Suitable for SMALLER R

P.86As Voltmeter

P.86As Ammeter

P.86As Ohmmeter

P.86

S is open

2

RR

RVV R

By division of voltage

A

C

R

S R

V

ε

A

C

R

S R

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.87

60 Ω resistors are neglected.

1010

R

RVR

V

10020

80

R

R 80R

10 Ω resistors are neglected.

6060

R

RVR

V

10012080

80

RV 40RV

P.87

V By division of voltage,

1 Ω resistors are neglected.

222

242 V 12

A

By division of voltage,

4.2122

22)2)(2(

2

22)2)(2(

2

V

)11(2 IV

2.1I

P.87

P.87

P.87

P.88

P.87

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 VB = VD

By division of voltage,

Voltage ratio in ABC = Voltage ratio in ADC

S

R

Q

P

P.89

ACVII )5.0)((2 21

PIVVV BAAB 1

RIVVV DAAD 2

QIIVVVV GBCBBC )( 1

SIIVVVV GDCDDC )( 2

GBDDB IVVV 5

P.89

At steady state, the capacitor is fully chargedVC = VB – VA

No current flowing through capacitor

By division of voltage,

2636

32

VVA

4621

24

VVB

Q = C(VB – VA)

Q = (20x10-6)(4 – 2)

Q = 40x10-6

QA = -40x10-6

Lower in potential

P.89 Metre Bridge

Metre bridge – simple form of Wheatstone bridge

P.89 Metre Bridge

If the reading of galvanometer is ZERO,

S

R

Q

P

P.90

When the bridge is balanced,

r

r

Q

P

S

R

2

1

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 l1 and l2 are of the same order.

2

1

S

R

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

P.90 Potentiometer

If the reading of galvanometer is ZERO,

V1 = VXY = VAC

(1) Protect galvanometer when large current flowing through it.

(2) Close the switch to have more accurate reading

P.91 Measurement of p.d.

l

l

V

V 11

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

l

lV 1

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

l

lV 1

Ammeter reading should be the same, otherwise

Adjust the value of rheostat until the reading of galvanometer is zero

R

VI

P.93 Measurement of internal resistance of a cell

Read the reading of point C when the switch is open.

VL

l0

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.

VL

lIRIr

0l

lIr

)( rRI

0l

l

rR

R

00

111

lRl

r

l

Plot a graph of (1/l) against (1/R)

P.93 Measurement of internal resistance of a cell

R

rR

r

10,01

When

10

rR

11

intercept on 1/R- axis

P.94

P.94

P.94

P.94