H H+ H
H2
Action of the cell
H+1
Zn (in solution) leave on zinc-
Zn - 2e Zn** (in solution) The hydrogen formed surrounds the anode.
1 is from copper.
copper sulphate Zn - 26. solution out-
side the porous cutt + 2e.
(at the Cu
cathode)
2NH + H2
Cathode Electrolyte
Depolamzer
dil. H2SO4
Cell
Anode
simple
cell
copper
zinc
頁二第張五第日四廿月二年癸夏
1973英文中學會考試題預習專欄
MAXAMI
物理科
(廿三)
Physics (23)
1. Cells & batteri es
WAH KIU YAT PO
報日僑華
rodes consist of a rigid frame work made of antimony-lead alloy into which the active elements is pasted under pressure. The active element is mainly lead oxide. The positive plates are turned into lead peroxide and negative plates. are turned into spongy lead. The electrolyte is dil. sulphuric acid The action of the cell is
Fb02 + Pb + H2SO
(charged)
三期星
Let the shunt be S. When the meter is fully deflected, the current through it is 5 mA. remaining current mist pass through the shunt. The current through the shunt is 1 0.005 – 0.995 A. The terminal voltage across the meter and the shunt is 0.1 W Hence
Se
0.1 0.995
0.1.005.
H2 formed surrounds the anode.
2e
Zn** (at the
cathode)
carbon
zine
pod
ride solution
surrounding
zine rod.
ammonium chlo-manganese
2NH4
dioxide + powd-
zinc rod.
the porous pot.
surrounding the carbon rod ing ered carbon
Daniel.
cell
copper
zine.
zinc sulphate
solution in
porous pot.
surrounding the pot.
Leclanche
cell
日八廿月三年三七九一层公年二十六國民華中育教僑華
50
15 resistance.
Then
is the current throug
By Ohn's law,
15-1
(30 – 1)10
ör 251
A..
2PbS0
2H20 (discharged)
The capacity of an accumulator is the amount of electricity which it can store. It is expressed in ampere-hours, A capacity of so ampere-hours means that the cell. can supply 4 ampere of current. Sor 20 hours; or 2 ampere of jurrent for 40 hours etc. she accumulator is constant at about 2.2 V.
4. Measuring instruments.
EMF of
(1) Conversion of a milli ammeter into a voltmeter. An ammeter or a milliammeter is an instrument for measuring the magnitude of current in the circuit. A volt- meter is an instrument used to. measure the potential difference. across resistance(s). Consider a moving coil instrument (the theory of which will be discussed later) which requires 5 mA for full scale deflection (f.s.d.)
Suppose the resistnace of the instrument is 20.
Then when
it is full scale deflected, the terminal voltage is
V = ir = 5 x 103 x 20 = 0,1
Since the coil obeys Ohm's law. the current thro' it is proport ional to the potential difference across it. On the other hand, the deflection is proportional to the current. It therefore follows that the deflection is proportions al to the potential difference. Thus the instrument can be used as a voltmeter giving reading up to 0.1 V.
20.2
MA
.e. a resistance of 0.1005 connected in paralled to the meter enables the meter to give current. up to 1 A.
5. Use of ammeter and voltmeter.
To measure the current through a resistance or drawn from the cell, an ammeter is connected in series with the resistance of the cell. Since the internal resist- ance of an ammeter is generally very small, the addition of an ammeter does not affect the current to a large extent. To measure the p.d. across a certain device, a voltmeter is connected in parallel with the device. The voltmeter is of very high resist- ance so that it takes only a negligible current and most of the current passes thro' the device. The reading of the volt- meter then gives the p.d. & across the device. In the case when a high resistance is required to be determined, a circuit shown in fig 4 is generally employed.
W
The ammeter A directly gives the current through the high resistan- ce R and the voltmeter reading is the sum of potential drops across R & the ammeter with a suitable correction, the high resistance R can be determined. Note that the circuit in fig 4 is essential when a high resistance is to be, determined. If a low resistance. is required to be determine, We employ the circuit shown in fig
Solution to exercise 2.
1. A bulb is rated at 120 V. 60W.
(a) The current it takes is
Current thro' ad
50-20
10 A
This is the current thro' ab,
ab and bd are in series, current thro' the branch bd
is also 10 A.
the
By a similar method as in finding the current fhro ab, we have 10i' = 10(10-1⋅ )
where is the currer
10: resistance.
11 = 5 A.
bc is.
Thus the current thro The voltage drop across be
x 5 25 V.
The eq resistance of the network is
R
R
The voltage drop across the parallel resistances is
RR
IR
The effective resistance of the circuit is
R 20
24
6.0
The
ent
the
ន
2
60
The current in the 16 & 201 resistances is 2 A since they are in series.
The voltage drop in the 16 resistance is 16 x 2 = 32 V. The voltage drop in the 201 resistance is 20 x 2 = 40 V. The voltage drop across the parallel resistances is 120 - 40 - 3248 VS. Applying the result in no. The current in the 60 ance is
..40
60 + 40
resist.
20.8 A.
The current in the 10n & the 30Л resistance is 2 - 0.8
1.2. A.
The voltage drop in the 100 resistance is 1.2 x 10 = 12 V. The voltage drop in the 301 resistance is
30 x 1.2 = 36. V.
6. The resistance of the lamp is
2. Defects of a simple cell:
(1) Local action.
Whenever two dissimilar 'metals are immersed in a chemical
solution, an emf is produced chemically. If the zinc contains impurities, small cells are formed between the zinc and its impuriti- es. This action does not add to the voltage of the cell, but it does use up zino. This is called the local action. To reduce this, the zinc is amalgamated with mercury so as to present a pure surface to the electrolyte.
(2) polarization:" The hydrogen formed tends to gather round the anode increasing the internal resistance of the cell. This effect is known as polarization. In a Leclanche's cell, this polarization is eliminated by using a depolarizer - manganese dioxide. The depolarizing action
is: 2Mnu2 + H2 M203+ However, the depolarization is not complete because Mm02 is a slow depolarizer. All the hydro- gen formed is not absorbed by the manganese dioxide and the cell polarizes. If the cell is allowed to stand idle for a period of tims the manganese dioxide will gradua- 1ly catch up, with the excess of hydrogen and depolarize the cell.
3. EMF of the cells.
(1) Simple cell:
slightly less than 1
(2) Daniel cell:/
about 1.08 V..
(3) Leclanche' cell:
about 1.5 V,
The 3 types of cells described above are called primary cells as they cannot be recharged when they are exhausted. A secondary cell is one which can be recharged when its life is exhausted. A common type of the secondary cell is the Lead-acid accumulator. The elect-
If the instrument is used to measure p.d. up to 1 v.
resistance is added.
a multiplier.
a series is called
Let it be R. When the meter is fully deflected, the current through it is 5 mA and is the.. same through R. The termal: voltage across the meter is 0.1 V. The voltage across R is 1 0.1
0.9 V. The value of R
0.9
5 x 10
1800
i.e. a resistance of 180 in series with the meter enables the meter to measure p.d. up to # V.
(2) Conversion of a milliammeter to an ammeter.
If the meter is used to measure current up to 1A, a parallel resistance usually called the shunt is used..
The
.60
A.
resistance..
120x120.
240 Ohma
The resistance of the wire of length 1 & of cross-section a is
R = p2 = 8√ where p is the
resistivity.
When the length is doubled and the diameter is halved the resistance is
100 x 100
·50
As shown in the cirquit
$1.00 0.5
=200
200
(a) The eq resistance of the cir-
cuit is
50.n
100
(b) The current thro' the lamp is
R
+200 200
50
00
ZA
(2 l)
та
By
X
645
The current thro' the 40 П the 60 resistance is
1.00
The
od
resistance between
1 A.
100
باکری
TO TO
The eq. resistance between bd is also 5
The eq. resistance between
The total current in the
circuit is.
글.
1 + 2 + 2 = 2 A.
(c) The power dissipated in the
200
•
ad =
6 J
10
+
15
A.
Tot. current in the circuit
100
Let i be the current thro' the
40
resistance
.1
x 200 = 50 W.
The power dissipated in the
& the 60 resistance is
r = 1 x 100 100 W.
( 未羌修入第五張第三言)