PHYSICS S.S ONE(EMF OF CELL)

SECOND TERM

WEEK: 8

ELECTROMOTIVE FORCE (EMF) OF CELLS

Cells

Cell is an electrical element where electrical potential is produced due to chemical reaction. Every electrochemical reaction has its limit of producing electric potential difference between two electrodes.
Cells are those where these electro-chemical reactions take place to produce the limited electric potential difference. For achieving desired electric potential difference across the battery terminals multiple numbers of cells are to be connected in series. Hence it can be concluded like that, a battery is a combination of several cells where a cell is a unit of a battery. For example, Nickel-cadmium battery cells normally develop about 1.2 V per cell while lead acid battery develop about 2 V per cell. So a 12 volt battery will have total 6 number of cells connected in series.

EMF of a cell

If anyone just measures the electric potential difference between two terminals of a battery when load is not connected with the battery, he or she will get the voltage developed in the battery when there is no current flowing through it. This voltage is generally referred as electromotive force or emf of battery. It is also referred as no-load voltage of battery.

Terminal Voltage of cell

Terminal voltage of cell is the potential difference across its terminals when the current is being drawn from it. Actually when load is connected with the battery, there will be load current flowing through it. As a battery is an electrical equipment, it must have some electrical resistance inside it. Because of this internal resistance of battery , there will be some voltage drops across it. So, if any one measures the terminal voltage of the load i.e. terminal voltage of cell when load is connected, he or she will get the voltage which is less than emf of the battery by internal voltage drop of the battery.

If E is the emf or no-load voltage of the battery and V is the terminal voltage of load voltage of the battery, then E – V = internal voltage drop of the battery.
As per Ohm’s law, this internal voltage drop is nothing but the product of electrical resistance offered by the battery and the current flows through it.

Internal Resistance of cell.

The entire resistance encountered by a current as if it flows through a battery from the negative terminal to the positive terminal is known as internal resistance of cell.

The electromotive force of the above circuit is represented by the formula  Where, r – internal resistance of the circuit. R – External resistance of the circuit. E – electromotive force. I – current.

The E.M.F of a cell is defined as the p.d across its terminals when it is in an open circuit i.e. not supplying current to an external circuit.

                 The internal resistance  ( r ) of a cell: is the opposition to current flow offered by the chemicals between the poles of the cell.

                 Lost p.d (Ir): is the potential drop across the internal resistance.

                 Terminal p.d is defined as the p.d between the terminals of a cell when it is delivering current to an external circuit.

CELLS IN SERIES AND PARALLEL

Cells in Series Connection

In series, cells are joined end to end so that the same current flows through each cell.  In case if the cells are connected in series the emf of the battery is connected to the sum of the emf of the individual cells. Suppose we have multiple cells and they are arranged in such a way that the positive terminal of one cell is connected to the negative terminal of the another and then again the negative terminal is connected to the positive terminal and so on, then we can that the cell is connected in series.

Equivalent EMF/Resistance of Cells in Series

If E is the overall emf of the battery combined with n number cells and E1, E2, E3 , En are the emfs of individual cells.

Then   E1 + E2 +  E3 + …….En 

Similarly, if r1, r2, r3, rn are the internal resistances of individual cells, then the internal resistance of the battery will be equal to the sum of the internal resistance of the individual cells i.e.

r = r1 + r2+  r3 +  rn

Cells in Parallel Connection

Cells are in parallel combination if the current is divided among various cells. In a parallel combination, all the positive terminal are connected together and all the negative terminal are connected together.

Equivalent EMF/Resistance of Cells in Parallel

If emf of each cell is identical, then the emf of the battery combined with n numbers of cells connected in parallel is equal to the emf of each cell. The resultant internal resistance of the combination is,

r =  (1/r1 + 1/r2 +  1/r3 +……..  1/rn )-1

Equivalent EMF/Resistance of Cells in Series and Parallel

Example

1.  A cell unknown e.m.f E and internal resistance 5Ω is connected to a 12Ω resistance. If the terminal p.d is 10v. calculate e.m.f.

Solution

 E=?; r=5Ω; R= 12Ω ; V=10v

E=I(R+r); E= V/R(R+r) = 10/12(12+5) =0.833(17) = 14.17v

Q2)  The p.d. across the terminals of a cell is 3.0 volts when it is not connected to a circuit and no current is flowing. When the cell is connected to a circuit and a current of  0.37 A is flowing the terminal p.d. falls to 2.8 V. What is the internal resistance of the cell?

 

1.      A voltmeter is connected in parallel with a variable resistance R which is in series with an ammeter and a cell. The cell is of e.m.f E and internal resistance r. For one value R1 of the variable resistance, the ammeter reads 0.3A and the voltmeter reads 0.9v. for another value R2 of the variable resistance, the ammeter reads 0.25A and the voltmeter reads 1.0v. neglecting the resistances of the meters. Find the values of R1, R2, E and r.

Solution

R1=?   I1= 0.3A V1= 0.9v   R2= ?  I2= 0.25A   V2= 1.0v

R1 = V1/ I1 = 0.9v / 0.3A = 3.0Ω ; R2 = V2 / I2 = 1.0v / 0.25A = 4.0Ω

E = V + Ir

E = 0.9v + 0.3r ; E = 1.0v + 0.25r equate the two equations

0.9 + 0.3r = 1.0 + 0.25r ;  0.3r – 0.25r = 1.0 -0.9 ; 0.05r = 0.1 ; r = 0.1 / 0.05 = 2Ω

E = 0.9 + 0.3r ; E = 0.9 + 0.3(2) ; E = 0.9 + 0.6 ; E = 1.5v

PRESENTATION

Step I: The teacher explains e.m.f ,  terminal p.d and internal resistance.

Step II: The teacher explains cells in series and parallel arrangement.

Step III: The teacher leads the students to solve the mathematical problems on e.m.f of a cell.

EVALUATION

The teacher evaluates the students by asking the following questions:

i.                     Define electromotive force

ii.                   Explain the emf and the r of a cell in series and parallel

iii.                  Solve mathematical problems on emf of cell.

 ASSIGNMENT

Explain why the e.m.f of a cell is greater than p.d across the cell when it is passing a current through an external resistance. A cell of e.m.f E and internal resistance r was connected in series with two external resistors A and B, of 4Ω and 2.5Ω respectively. When a high resistance voltmeter was connected across A, the reading was found to be 2v. when another  resistor of 4 Ω was connected across  A and B, the voltmeter reads 3v. Draw circuit diagrams of two arrangements. Calculate the e.m.f and the internal resistance of the cell.`

  

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