Current Electricity
Electric Current and Ohm’s Law Elementary Idea Of Secondary Cell Or Storage Cell
1. lead-add Accumulator: it is a secondary ceil. French Gaston Plante was discovered in 1859.
Discussion:
Positive electrode: The active element of this electrode is lead dioxide (PbO2).
Negative electrode: The active element of this electrode is metallic lead, taken in a spongy form.
Class 12 Physics Ohm’s Law Solutions
Electrolyte:
Dilute sulphuric acid of specific gravity 1.25 is taken as an electrolyte In a thick glass or bakelite vessel. The plates are dipped in the acid. Only the terminals of the two electrodes remain outside the vessel
The electromotive force of the cell:
When a fully charged cell begins to discharge, Its emf is 2.2 V. But after a short while the emf comes down to 2,0 V and remains constant for a long time. At last, when the cell has fully discharged the emf the cell comes down to about 1.8 V. To ascertain whether the cell has been fully charged and Is in an active state, or has been folly discharged, a voltmeterInsertedin the external circuit can well serve the purpose. The ingredients of the cell need not be Inspected.
It Is to be noted that during the time of discharge, lead sulfate Is formed at the two electrodes, and the active elements i.e., Pb, PbO2 and H2SO4 gradually decay.
The specific gravity of sulphuric acid:
The cell contains sulphuric acid of a specific gravity of 1.25. During discharging, the specific gravity of the sulphuric acid solution decreases. When the cell is fully discharged the specific gravity of the add solution falls to 1.18. During charging, sulphuric add is regenerated.
Class 12 Physics Ohm’s Law Solutions
When the cell is fully charged the specific gravity of sulphuric acid comes back to Its normal value of 1 .25. So, even by measuring the specific gravity of sulphuric acid, the condition of the cell can be determined.
However, to determine the condition of the cell, measurement of the emf of the cell with a multimeter is a better option.
Sulphuric acid Use:
The internal resistance of the cell is very low i.e., current flows through the cell almost without any resistance. So this cell is used for getting a steady current of high magnitude in the external fruit for a long time.
It is used O in cars, buses, trains, and even in laboratories, and Q in inverters, for generating electricity in houses during power cuts.
2. Alkali Accumulator:
The active elements of this cell are iron (negative plate), nickel hydroxide (positive plate), and caustic potash solution (KOH) (electrolyte). This is also called a nickel-iron accumulator or nife cell. This cell is also known as Edison cell after the name of its discoverer Thomas Alva Edison
Alkali Accumulator Disadvantages:
In comparison to lead-acid accumulator:
- Its internal resistance is high.
- Emf is low (1.3 V).
- Efficiency is small
Alkali Accumulator Advantages:
- No internal disturbance occurs on heavy jerking.
- It can be left for a long time in a fully charged or in a fully discharged condition.
- No damage is done if it is overcharged or over-discharged.
Capacity and Efficiency erf a Secondary Gell:
Capacity:
The capacity of a secondary cell is defined as the amount of electricity (charge) that the cell is capable of supplying in the external circuit before being completely discharged
Unit of capacity: Ampere-hour (A-h) is the unit of capacity.
1 A.h = 1 A x 1 h
= 1 A x 3600 s
= 3600 A.s
= 3600 C
So, a ceiling having a capacity of 1 A.h can supply a charge of 3600 C. Obviously A.h is a big unit.
Example:
By the statement that the capacity of a secondary cell is 50 A – h, we mean that the cell can supply a current of 1 amperes for 50 hours or 2 amperes for 25 hours. However, the capacity expressed in ampere-hour is always an approximate value only.
Class 12 Physics Solved Examples Ohm’S Law
Efficiency:
The amount of charge given to a secondary cell during charging cannot be back totally during discharging. The ratio of the charge obtained to the amount given is called ampere-hour efficiency. In the case of lead-acid accumulator is 0.9 or 90%.
Again the amount of external energy supplied to a secondary cell during charging cannot be get back totally in the form of electrical energy during discharging. The fraction of the supplied energy obtained from the cell is called the energy efficiency or watt-hour efficiency of the cell.It is given by,
⇒ \(\eta=\frac{\text { energy obtained during discharging }}{\text { energy supplied during charging }}\)
⇒ \(\frac{average emf during discharging x amount of charge obtained}{average emf during charging x amount of charge supplied}\)
⇒ \(\frac{\text { average emf during discharging }}{\text { average emf during charging }}\) x ampere-hour efficiency
During charging of a lead-acid accumulator the external source used has an average emf of 2.2 V. But during discharging the average emf obtained from the accumulator is 2.0 V. So, the energy efficiency of a lead-acid accumulator is,
⇒ \(\eta=\frac{2.0}{2.2} \times 0.9=0.8=80 \%\)
Current Electricity
Electric Current and Ohm’s Law Elementary Idea Of Secondary Cell Or Storage Cell Numerical Examples
Example 1. A battery Is charged at a potential of 15 V for 8 h by means of a current of 10 A. While discharging it supplies a current 5 A for 15 h at a potential difference of 14 V, Calculate the watt-hour efficiency of the battery.
Solution:
Watt-hour efficiency or energy efficiency
⇒ \(=\frac{\text { energy obtained during discharging }}{\text { energy supplied during charging }}\)
⇒ \(\frac{14 \times 5 \times 15}{15 \times 10 \times 8}\)
= \(\frac{7}{8}\)
= 0.875
= 87.5%
The watt-hour efficiency of the battery = 87.5%
Differences between Primary and Secondary Cells:
