Internal resistance of battery

Document Sample

```					        Internal resistance of battery
Some of the electrical energy is
dissipated by Joule heating inside the
battery.
e.m.f. across the + & - terminal              ξ
of the battery is lower than the
marked value when connected to                     r
external components.

Terminal voltage V
∴ 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.
Finding the internal resistance of the battery
The current I through the circuit is varied
by a resistance box which has known value
R
of resistance R.
ξ= V + I r      =IR+Ir                  ξ
I                        I
=> ξ = R + r                                       r
I
=> R = ξ        - r
V
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.
Combination of Resistors
1. 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
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.

I  I1  I 2  I 3      
V
R1

V
R2

V
R3
 1
V
1
R  R R 
 1   2
1 

3

1   1   1   1
      
R   R1 R2   R3
Power and heating effect
The charge pass through the resistor:
Q  I  t
The electrical energy converted into
other form of energy in Δt:

W  V  Q
∴ Power of the resistor:

P
W V  Q
t

t
V 
Q
t              V  I
V2                              2
 P  V  I  I 2R            P    I  I 2R 
R                              R
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.
V2               V 2 12 2
P               R          14 .4
R                P   10
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

2                2
P  I  R  0.619 14 .4  5.5W
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.

V2              V 2 92
P              R        1.62 
R               P 50

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
2              2
P  I  R  0.775 11 .62   6.97W
Power output and Resistance

The current flow through the above circuit:                  
I
Rr
Power output through R:                2                  2R
Pout  I R  R  r 2
 2r            2r         2
Max power occurs when R = r      Pmax                               
r  r    2
4r   2       4r
Proof of max. Power output

 2R                       R
2
P                  
R  r 2              2
R  2 Rr  r   2

 R 2

2                    2
R  2 Rr  r  4 Rr
 R 2
d  R 
                                         Try             2
dR  R  r  
2
( R  r )  4 Rr                                       
Efficiency of Electric Circuit

In general the power of the circuit.
Power output      I 2R    R
                              100%
Power input   I (R  r) R  r
2

When max. power occurs, R = r, the efficiency is:

r
     100%  50%
rr
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.
2     1.52
 Po    I  I 2 r           1.125W
r     2
b.)    Max. useful power output occur when R = r
i.e. external resistance = 2Ω
21.52
 Pmax            0.28W
4r 4  2
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.
2   92
 Po    I  I 2 r         16 .2W
r    5
b.)   Max. useful power output occur when R = r
i.e. external resistance = 5Ω
2 92
 Pmax            4.05W
4r 4  5
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          2
Ploss  I L  R

∵ Power transfer to the user equal to the
power generate, it usually remains constant
Typical Example ---- Combination of 2 cells

ξ   r1
Since any charge + or – cannot
+                -
climb through the energy
barrier set by the other cell,              ξ   r2
NO current flow through the
cells.
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

Net e.m.f. = 14 V – 12 V = 2 V
20Ω
Total resistance = 15 + 5 + 20 = 40Ω
12V    15Ω
Current = 2 V / 40 Ω = 0.05 = 50 mA
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.
9V      10Ω

Net e.m.f. = 9 V – 1.5 V = 7.5 V
50Ω
Total resistance = 10 + 5 + 50 = 65Ω
1.5V 5Ω
Current = 7.5 V / 65 Ω = 0.115 A
2 parallel cell connected to a resistor
I1 ξ   r1
The terminal voltage of the cells
are the same and equal to the
voltage across R                              I2 ξ   r2

ξ = I1 r1 + I R
ξ = I2 r2 + I R
I1 + I2 = I                            I          R

This is known as Kirchoff’s
1st law of current
Example 6 The light bulb in the diagram shown below has
a resistance of 6 Ω. Find the power output.
1.5 V = I × 6Ω + I1 × 2Ω --- ( 1 )            I1 1.5 V
2Ω
1.5 V = I × 6Ω + I2 × 4Ω --- ( 2 )
I = I1 + I2 --- ( 3 )                     I2 1.5 V
1.5 V = (I1 + I2 ) × 6Ω + I1 × 2Ω                       4Ω
1.5 V = (I1 + I2 ) × 6Ω + I2 × 4Ω
I        6Ω
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
Classwork 4 The light bulb in the diagram shown
below has a resistance of 10 Ω.
Find the power output.
I1 1.5 V
1.5 V = I × 10Ω + I1 × 5Ω --- ( 1 )                   5Ω
1.5 V = I × 10Ω + I2 × 2Ω --- ( 2 )
I2 1.5 V
( 1 ) × 2, ( 2 ) × 5                                    2Ω

I = 0.13125 A                         I        10Ω

∴ Power output of the bulb = 0.131252 × 10 = 0.17226 W
I-V characteristics of junction diode ( semiconductor diode )

A diode allow current flow in 1         I / mA
direction only. However, the
voltage across the diode must reach
a certain value to enable the charge 20 mA
VD / V
carriers to flow.
0.8 V
For the diode to operate, I ≧20 mA,
the largest I                             0.8 V      2.2 V
VR
I min          20 mA
Rmax
R
VR    2.2
Rmax                110 
I max 20 mA                        I
ξ= 3V
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:

For the diode to operate safely,
Pmax  VD  I max
0.8 V     2.2 V
0.08  0.8V  I max  I max  0.1A

VR    2 .2
Rmin              22 
I max 0.1A                                    R

∴ the diode can be operated with         I
ξ= 3V
a resistor of 22Ω≦R ≦110 Ω
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.
I / mA                          1.2 V      7.8 V

R
50 mA               VD / V
1.2 V                    I
ξ= 9V

Rmax = 156 Ω                     Rmin = 9.36 Ω
Thermionic diode 熱放電二極管
I / mA
A thermionic diode is a evacuated
glass tube with a heated cathode.   2 mA
Electrons is excited in the cathode
and ‘jump’ to the anode inside the                   VD / V
glass tube.
4V
If a diode is saturated at I = 2 mA,
VD         12V- VD
the number electrons reaches a
maximum value and nor more
electrons can flow inside the glass      -
-          R
tube.

I
ξ= 12V
I / mA                              VD       12V- VD

2 mA
-
-         R
VD / V

4V                             I
ξ= 12V
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
Bridge circuit

1Ω    1Ω        1Ω      1Ω
No current
∵ same p.d.

10Ω   10Ω       10Ω    10Ω

4V
P          Q
I1

I2

R          S

VP= VR     I1 P = I2 R   P R

VQ= VS     I1 Q = I2 S   Q S
Classwork 6 The ammeter in the diagram below shows
no reading. Calculate the resistance of the unknown
resistor P

P           25Ω

64Ω            40Ω

P = 40 Ω

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 139 posted: 12/30/2010 language: English pages: 27
Description: Internal resistance of battery