AL Current Electricity P.66. P.69 Conduction electrons collide the lattice ions only due to higher...

AL Current Electricity

P.66

P.69

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

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.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.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.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.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


6060

R

RVR

V

10012080

80

RV 40RV

P.87

V By division of voltage,


222

242 V 12

A


4.2122

22)2)(2(

2

22)2)(2(

2

V

)11(2 IV

2.1I

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


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


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


S

R

Q

P

P.90 Potentiometer


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)

AL Current Electricity P.66. P.69 Conduction electrons collide the lattice ions only due to higher...

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Transcript of AL Current Electricity P.66. P.69 Conduction electrons collide the lattice ions only due to higher...