Electromotive Force (E.M.F.):
Electric current flows through a conductor due to the motion of charge. We know that motion cannot take place without the application of force. So we can safely, assume that the flow of current is the result of application of a force on the charge carriers. This force comes into existence only when a source of electricity (a cell, battery, etc.) is connected across the conductor. Due to some process in source of electricity, energy is liberated which crates a potential difference across the two ends of conductor. As a result of this an electric field is set up in the conductor which exerts a force on the charge, thus, producing motion on it. As the charge 'q' is taken round the circuit, some work 'W' is done.
Work done per unit charge to take the charge around the circuit is called electromotive force.
E.M.F = W/q
Unit of E.M.F. in S.I. is joule/coulomb.
Since joule/coulomb represents the potential difference so, potential difference and E.M.F. are dimensionally similar.
Electro-motive force (e.m.f.)
It is the difference of potential across the two terminals of a source of current when there is no electric current flowing through the external resistance.
Electromotive Force and Potential Difference:
Fig. 1. shows a source of current i.e., a cell connected to a conductor of resistance R through a key K. A voltmeter V is connected across the two terminals of the cell.
When the key K is open [Fig. 1(i)], the reading of voltmeter gives the electromotive force (e.m.f.) E of the cell.
The electromotive force E of a cell is defined as the difference of potential between its terminals when there is no current in the external circuit, i.e., when the cell is in open circuit.
When the key K is open [Fig. 1(ii)], an electric current i flies from positive to negative terminal in the external circuit and from negative terminal to positive terminal within the cell. Thus, a potential drop 'r' takes place across the internal resistance 'r' (shown separately) of the cell. The direction of 'ir' is opposite to the e.m.f. 'E' of the cell. The reading of the voltmeter now decreases and gives the potential difference V between the terminals of the cell.
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| Fig. 1. Distinction between e.m.f. and potential difference. |
The potential difference of a cell is the difference of potential between two terminals when it is in closed circuit.
∴ V = E - ir or E = V + ir
Hence, E is greater than V by an amount equal to the drop of potential within the cell due to its internal resistance.
Distinction Between E.M.F. (E) and Potential Difference (V):
E.M.F. (E) and potential difference 'V' can be distinguished from each other in following respects.
1. E.M.F. is the cause while potential difference is its effect.
2. It is associated only with cells while potential difference can exist across a resistor.
3. E.M.F. is measured by the motion of a charge around a circuit while potential difference is measured by motion of charge between two points.
4. E.M.F. is measured in an open electric circuit while potential difference measured in a closed electric circuit.
5. Refer Fig. 2(i), cell is getting discharged through an external resistance R.
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| Fig. 2. Discharging and charging of a cell. |
i = E/R + r or E = iR + ir
Where 'r' is 5he internal resistance of the cell.
Since, iR = potential difference across external resistance = V
∴ E = V + ir or E - V = ir .... (i)
During charging Fig. 2(ii), the charger sends a current through the cell. It can be noted that direction of current through internal resistance 'r' is in direction opposite to that in first case.
Charge i to -i in equation
E - V = -ir .... (ii)
From equations (i) and (ii) we can conclude that :
- E > V during discharging of cell
- E > V during charging of cell
- E = V in open circuit (i = 0).
Examples-
Solution-
Using equation (ii), we get n = 12
Thus, the best mode of grouping will be to have 12 cells connected in series with each other and 3 cells such rows in parallel.
Current i = mnE/mR + nr
= 36 × 1.5/3 × 2 +12 × 0.5 . A
= 36 × 1.5/12
= 4.5 A.
2) 12 cells each having the same e.m.f. are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is in series with an ammeter and two cells identical with other. The current is 3 A when the external cells aid this battery and is 2 A when the cells oppose the battery. How many cells in the battery are wrongly connected ?
Solution-
Let 'E' be the e.m.f. of each cell and 'n' be number of cells wrongly connected.
Number of cells in correct order = 12 - n
Net e.m.f. of the battery
= (12 - n) E - nE
= (12 - 2n) E
(i) When the two external cells aid this battery
Net e.m.f. = (12 - 2n) E +2E
∴ (12 - 2n) E +2E/R = 3
where R = external resistance.
or (14 - 2n) E = 3R ... (i)
(ii) When the two external cells oppose this battery
Net e.m.f. = (12 - 2n) E - 2E
∴ (12 - 2n) E - 2E/R = 2
or (10 - 2n) E = 2R ... (ii)
Dividing equation (i) by equation (ii),
(14 - 2n) E/(10 - 2n) E = 3R/2R
28 - 4n = 30 - 6n
or 2n = 2
or n = 1
Thus, there is only one cell which is wrongly connected in the battery.
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