# Concept of EMF , Terminal Potential Difference

The natural tendency of a positively charged particle is to move from higher potential to lower potential. If a current is to flow through a conductor ‘R’, there must be a potential difference across its ends A and B. But in order to establish current in a closed circuit it is necessary to do continuous work on electric charge flowing through it.

$\displaystyle \oint \vec{E}.\vec{dl} \ne 0$ (for a closed loop) unlike in electrostatics.

This condition, when established in a closed circuit, is known as electromotive force (emf).

## E.M.F :

The work done by the electric field  on a unit positive charge when it is taken through a closed loop is known as emf (acting within loop).
This is usually obtained by connecting a source of E.M.F. (e.g. a cell) across the conductor. The source maintains a constant potential difference across the ends A and B. Current will flow from A to B if the end ‘A’ is connected to the positive terminal (higher potential) and B to negative terminal (lower potential)

This source of EMF is like a pump which keeps on transferring the charge from B to A through itself and hence it has to perform work (as it transfers the negative charge from low to high potential) If this work done in transferring change q is W, then by the definition, the E.M.F. of the battery is

$\large E = \frac{W}{q}$

The EMF of a cell (and that of any other source of electrical energy) is also equal to the energy converted into electrical energy from other forms (e.g. chemical, mechanical) when unit charge passes through it.
The chemicals within cell offer resistance to current flow, also known as the internal resistance of the cell.

## Terminal Voltage:

When no current flows through the cell, the circuit is said to be an open circuit.

In such a case, the potential difference (p.d) across the terminals of the cell, called the terminal voltage (V) will be equal to the emf (E) of the cell.
If an external resistance R is connected across the two terminals of the cell, as in figure then current flows in the closed circuit.,

$\large i = \frac{V}{R}$

and also , $\large i = \frac{E}{R+r}$

$\large i R + i r = E$

$\large V + i r = E$

$\large V = E – i r$

Lost volts: It is the difference between emf and P.D. of a cell it is used in driving the current between terminals of the cell. Lost volts E – V = i r

### Different Concepts with a Cell:

(i)When the cell is charging, the EMF is less than the terminal voltage (E < V) and the direction of current inside the cell is from +ve terminal to the –ve terminal.

$\Large V = E + i r$

(ii)When the cell is discharging, the EMF is greater than the terminal voltage (E > V) and the direction of current inside the cell is from ve terminal to the +ve terminal.

$\Large V = E – i r$

Hence ,  E > V

### Back EMF:

When current flows through the electrolyte solution, electrolysis takes place with a layer of hydrogen and this hinders the flow of current. In the neighborhood of both electrodes, the concentrations of ions get altered. This opposing EMF is called back EMF and the phenomenon is called Electrolytic polarisation. To reduce back emf manganese dioxide (or) potassium dichromate is added to electrolyte of cell.

Problem : Figure shows a source (a battery) with an emf E of 12 V with an internal resistance r of 2Ω and an external resistance of 4 Ω is added to complete the circuit. What are the voltmeter and ammeter readings .

Solution : An ideal voltmeter has infinite resistance, hence no current flows through it. An ideal ammeter has zero resistance .

The current i through the resistor R

$\large i = \frac{E}{R+r}$

$\large i = \frac{12}{4 + 2} = 2 A$

The ammeter will read , i = 2 A

As there is no potential difference between points a and p or between points b and q.

That is, Vab = Vpq

We can find Vab by considering a and b either as the terminals of the resistor or as the terminals of the source.

Vpq = iR = 2 × 4 = 8V

Considering them as the terminals of the source, we have

Vab = E − ir = 12 − 2 × 2 = 8V

Either way, we conclude that the voltmeter reads Vab = 8V

Problem :If in the previous problem the battery is being charged with a current 5 amp. in what direction will current flow inside the battery, and what will be the reading of the voltmeter ?

Solution: In charging a battery, positive terminal of the battery is connected to positive terminal of the charger. Hence the current inside the battery is from anode to cathode.
The net change in potential energy for a charge q making a round trip around a complete circuit must be zero.

Hence the net change in potential around the circuit must also be zero, in other words, the algebraic sum of potential difference and emf ‘s around the loop is zero.

Considering the section a b

Va − Vb = E + Ir

Vab = E + Ir

Vab = 12 + 5 × 2 = 22 volt

Note : If the terminals of the battery are connected directly to each other with no external resistance, then situation is called a short circuiting of battery.
Problem : In the circuit shown in the figure each battery is 5 V and has an internal resistance of 0.2 ohm. What is the reading of an ideal voltmeter V ?

Solution : As the internal resistance of an ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as

1/r1 = 1/r + 1/∞

=> r1 = r

Now as the given 8 batteries are discharging in series i.e.

Eeq = 8 × 5 = 40 V

and req = 8 ×0.2 = 1.6 Ω

So current in the circuit I = = 25 A
Hence potential difference across the required battery

V = E − Ir

= 5 − 25 × 0.2

= 0 V.

Problem  : Two cells, having emf. of 10 V and 8V respectively, are connected in series with a resistance of 24 Ω in the external circuit.  Their internal resistances are 2Ω/V (proportional to their emfs). Find the current in the circuit .

Solution: We determine the internal resistance of each these cells :

r1 = 2 × 10  = 20 Ω

r2 = 2 × 8 = 16 Ω

Total resistance in circuit = (24 + 16 + 20)Ω = 60 Ω

Current , $\displaystyle I = \frac{18}{60}$

I = 0.3 A