Docstoc

Internal resistance of battery

Document Sample
Internal resistance of battery Powered By Docstoc
					        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:
Stats:
views:139
posted:12/30/2010
language:English
pages:27
Description: Internal resistance of battery