al current electricity p.66. p.69 conduction electrons collide the lattice ions only due to higher...
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AL Current Electricity
P.66
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P.69
Conduction electrons collide the lattice ions only due to higher chance.
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P.74
73 Ω
Resistance Box
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P.74
Potential Divider
R2
R1
A
C
21
0
RR
VI
Vo
Vout
021
111 V
RR
RIRVVout
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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
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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
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P.75
nAq
Iv
2
1
Anq
I
dA
dv
When cross-sectional area A is increasing,
then the drift velocity is decreasing
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P.75v
nAq
Iv
A
lR
IRV
lvnqnqvlA
lIIRV )()(
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P.75
R1 = V / I = 1 Ω R2 = V / I = 2 / 5 = 0.4 Ω
ΔR = R2 - R1 = 0.4 – 1 = - 0.6 Ω
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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
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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
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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
PX = (0.24)(50) = 12 (W)
For Y, I = 0.24 A and V = 150 V
PX = (0.24)(150) = 36 (W)
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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
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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
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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
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
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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.
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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 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.
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P.79
)(
2
rR
VP
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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.80
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P.80
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P.80
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P.80
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P.81
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P.81
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P.81
Vx is equal to e.m.f. 2V
2)6(105
5
kk
kVX
X
No voltmeter reading
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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)
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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
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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'
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P.83
Draw most of voltage across R Draw most of current through S
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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
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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 + RM)
5 = (100μ)(Rtotal)
Rtotal= 50 kΩ
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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
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P.86As Voltmeter
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P.86As Ammeter
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P.86As Ohmmeter
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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
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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.
1010
R
RVR
V
10020
80
R
R 80R
10 Ω resistors are neglected.
6060
R
RVR
V
10012080
80
RV 40RV
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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
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P.87
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P.87
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P.87
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P.88
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P.87
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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 VB = VD
By division of voltage,
Voltage ratio in ABC = Voltage ratio in ADC
S
R
Q
P
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P.89
ACVII )5.0)((2 21
PIVVV BAAB 1
RIVVV DAAD 2
QIIVVVV GBCBBC )( 1
SIIVVVV GDCDDC )( 2
GBDDB IVVV 5
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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
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P.89 Metre Bridge
If the reading of galvanometer is ZERO,
S
R
Q
P
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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
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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
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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
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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
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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)
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P.94
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P.94
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