internal resistance of battery
DESCRIPTION
ξ. r. Internal resistance of battery. e.m.f. across the + & - terminal of the battery is lower than the marked value when connected to external components. Some of the electrical energy is dissipated by Joule heating inside the battery. Terminal voltage V. - PowerPoint PPT PresentationTRANSCRIPT
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Internal resistance of batterySome of the electrical energy is dissipated by Joule heating inside the battery.
r
ξ
Terminal voltage V
e.m.f. across the + & - terminal of the battery is lower than the marked value when connected to external components.
∴ the voltage across the terminal of a cell is called the terminal voltage and is usually less than the e.m.f. of the cell.
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Finding the internal resistance of the battery
The current I through the circuit is varied by a resistance box which has known value of resistance R.
I I
R
V
ξ= V + I r ξr
= I R + I r
=> ξ = R + r
I
=> R = ξ - r
I
∴ by plotting a graph of R against 1 / I, a straight line can be obtained. The slope of the graph is ξ and the intercept on the y-axis is the internal resistance r.
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Combination of Resistors1. In series
All the resistors carry same current I
V = V1 + V2 + V3
= I R1 + I R2 + I R3
= I ( R1 + R2 + R3 )
∴ R = R1 + R2 + R3
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Combination of resistors
2. In parallel
The p.d. V across each resistors is the same., but the current branches into I1, I2 and I3.
321 IIII 321 R
V
R
V
R
V
321
111
RRRV
321
1111
RRRR
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Power and heating effect
The charge pass through the resistor:
tIQ The electrical energy converted into other form of energy in Δt:
QVW
∴ Power of the resistor:
t
WP
t
QV
t
QV
IV
R
VRIIVP
22
RRIIP
22
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Example 5 A light bulb labelled 12 V 10W is connected across a 12 V cell with internal resistance 5Ω. Find the power output.
∵ the cell has internal resistance, the p.d. across the light bulb ≠12 V. But the resistance of the bulb is fixed.
4.1410
12222
P
VR
R
VP
The total resistance of the circuit is 5Ω + 14.4 Ω = 19.4 Ω
The current flow through the circuit and the bulb is = 12 V / 19.4 Ω = 0.619 A
WRIP 5.54.14619.0 22
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Classwork 1 A student connect a toy motor labelled 9 V 50 W across a 9 V battery with internal resistance 10 Ω. Find the power output by the toy motor.
62.150
9222
P
VR
R
VP
The total resistance of the circuit is 1.62Ω + 10 Ω = 11.62 Ω
The current flow through the circuit and the toy motor is = 9 V /11.62 Ω = 0.775 A
WRIP 97.662.11775.0 22
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The current flow through the above circuit:rR
I
Power output through R:RIPout
2 22
rR
R
Power output and Resistance
Max power occurs when R = r rr
r
rr
rP
44
2
2
2
2
2
max
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22
rR
RP
2
rR
R
dR
dTry
Proof of max. Power output
22
2
2 rRrR
R
RrrRrR
R
42 22
2
RrrR
R
4)( 2
2
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Efficiency of Electric Circuit
%100)(input
output 2
2
rR
R
rRI
RI
Power
Power
In general the power of the circuit.
When max. power occurs, R = r, the efficiency is:
%50%100
rr
r
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Example 6 A 1.5 V cell has an internal resistance of 2 Ω. Find the condition for a.) Max. Power.b.) Max. Efficiency.
a.) Max. Power consumption occur when the cell is short circuit.
Wr
rIIPo 125.12
5.1 222
b.) Max. useful power output occur when R = ri.e. external resistance = 2Ω
Wr
P 28.024
5.1
4
22
max
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Classwork2 A 9 V cell has an internal resistance of 5 Ω. Find the power out put for a.) Max. Power.b.) Max. Efficiency.
a.) Max. Power consumption occur when the cell is short circuit.
Wr
rIIPo 2.165
9222
b.) Max. useful power output occur when R = ri.e. external resistance = 5Ω
Wr
P 05.454
9
4
22
max
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Power transmission
In power transmission, the voltage across the power cable is VL – V’L. or ILR, where L is the current through the cable.
The power loss by the cable is
RIP Lloss 2
∵ Power transfer to the user equal to the power generate, it usually remains constant
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Typical Example ---- Combination of 2 cells
ξ r1
ξ r2
+ -Since any charge + or – cannot climb through the energy barrier set by the other cell,
NO current flow through the cells.
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Example 6 A 12V battery of internal resistance 15 Ω is recharged by a 14 V d.c. supply with internal resistance 5Ω via a 20 Ω. Find the current through the battery.
14V 5Ω
12V 15Ω
Net e.m.f. = 14 V – 12 V = 2 V
Total resistance = 15 + 5 + 20 = 40Ω
Current = 2 V / 40 Ω = 0.05 = 50 mA
20Ω
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Classwork3 A student uses a 9V battery of internal resistance 10 Ω to charged a 1.5 V ‘AA’battery with internal resistance
5Ω via a 50 Ω resistor. Find the current through the battery.
Net e.m.f. = 9 V – 1.5 V = 7.5 V
Total resistance = 10 + 5 + 50 = 65Ω
Current = 7.5 V / 65 Ω = 0.115 A
9V 10Ω
1.5V 5Ω
50Ω
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2 parallel cell connected to a resistor
ξ = I1 r1 + I R
ξ r1
ξ r2
I1
I2
I R
The terminal voltage of the cells are the same and equal to the voltage across R
ξ = I2 r2 + I R
I1 + I2 = I
This is known as Kirchoff’s 1st law of current
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Example 6 The light bulb in the diagram shown below has a resistance of 6 Ω. Find the power output.
1.5 V2Ω
1.5 V
4Ω
I1
I2
I 6Ω
1.5 V = I × 6Ω + I1 × 2Ω --- ( 1 )
1.5 V = I × 6Ω + I2 × 4Ω --- ( 2 )
I = I1 + I2 --- ( 3 )
1.5 V = (I1 + I2 ) × 6Ω + I1 × 2Ω
1.5 V = (I1 + I2 ) × 6Ω + I2 × 4Ω
Solve, I1 = 0.06818 A and I2 = 0.136 A
i.e. I = 0.06818 + 0.136 = 0.2042 A
∴ Power output of the bulb = 0.20422 × 6 = 0.25 W
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Classwork 4 The light bulb in the diagram shown below has a resistance of 10 Ω. Find the power output.
1.5 V 5Ω
1.5 V
2Ω
I1
I2
I 10Ω
1.5 V = I × 10Ω + I1 × 5Ω --- ( 1 )
1.5 V = I × 10Ω + I2 × 2Ω --- ( 2 )
( 1 ) × 2, ( 2 ) × 5
I = 0.13125 A
∴ Power output of the bulb = 0.131252 × 10 = 0.17226 W
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I-V characteristics of junction diode ( semiconductor diode )
ξ= 3V
R
I
0.8 V 2.2 V
A diode allow current flow in 1 direction only. However, the voltage across the diode must reach a certain value to enable the charge carriers to flow.
I / mA
VD / V20 mA
0.8 VFor the diode to operate, I 20 mA, ≧the largest I
mAR
VI R 20
maxmin
11020
2.2
maxmax mAI
VR R
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ξ= 3V
R
I
0.8 V 2.2 V
For the diode to operate safely,
maxmax IVP D
221.0
2.2
maxmin AI
VR R
For a diode to operate safely, prevent it burn out, a maximum current must be noticed. This current is limited by the power rating of the diode.Assume a diode has a maximum rating of 0.08 W, the smallest R is given by:
AIIV 1.08.008.0 maxmax
∴ the diode can be operated with a resistor of 22Ω R 110 Ω≦ ≦
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ξ= 9V
R
I
1.2 V 7.8 VI / mA
VD / V50 mA
1.2 V
Classwork 5 The I-V characteristic of a diode is shown below.The diode is operated with a resistor R. Given that the maximum power rating of the diode is 1 W. Calculate the maximum and minimum value of the resistor.
Rmin = 9.36 ΩRmax = 156 Ω
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Thermionic diode 熱放電二極管
A thermionic diode is a evacuated glass tube with a heated cathode. Electrons is excited in the cathode and ‘jump’ to the anode inside the glass tube.
I / mA
VD / V
2 mA
4 V
ξ= 12V
R
I
VD 12V- VD
- -
If a diode is saturated at I = 2 mA, the number electrons reaches a maximum value and nor more electrons can flow inside the glass tube.
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ξ= 12V
R
I
VD 12V- VD
- -
I / mA
VD / V
2 mA
4 V
If the diode is just saturated, the voltage across the diode VD is 4 V and the current I = 2 mA.
12 V = VD + I R = VD + 2 mA × R VD = 12 V - 2 mA × R
Since I must be larger than 2 mA, the resistor must be smaller than a certain value.
VD > 4V 12V – 2 mA × R > 4V
∴ R < 4000 Ω
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Bridge circuit
1Ω 1Ω 1Ω 1Ω
10Ω10Ω10Ω10Ω
No current same p.d.∵
4 V
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P Q
SR
VP= VR I1 P = I2 R
VQ= VS I1 Q = I2 S S
R
Q
P
I1
I2
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Classwork 6 The ammeter in the diagram below shows no reading. Calculate the resistance of the unknown resistor P
P 25Ω
40Ω64Ω
P = 40 Ω