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# Electromotive Force

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 Sub Topics Electromotive force (emf) is the potential difference generated in a circuit. One can relate emf with the flow of current. To know about the emf, let us explain a small example. If water is flowing from one end to other end, there should be a pressure difference between the both ends. If there is no difference in pressure, water will not flow. The given figure gives a clear idea about the above statement. This example is same as that of current flow. There is a force or potential difference is present in an electrical circuit, the current will flow from higher potential to lower potential. This potential difference or driving force is termed as electromotive force.

## Electromotive Force Definition

Electromotive force or emf can be defined as the potential difference or driving force which causes the flow of electricity through a circuit.

From the figure it is clear that, the battery provides the emf for this circuit. The current will flow from positive terminal to negative terminal through a resistor.

## Electromotive Force Equation

Electromotive force equation can be derived using the simple circuit which is given below:

From the figure, E is the emf of the battery, which is connected with a resistance R and r be the internal resistance of the battery. Total voltage in the terminal of  the battery be E - Ir .

We know that, terminal voltage of the battery is equal to the potential difference across R. So,
E - Ir = IR
E = IR + Ir
E = I (R + r)

I =$\frac{E}{R+r}$

## Back Electromotive Force

Back electromotive force is defined as the electromotive force generated in such a way that which pushes the current in opposite direction. It is also known as counter electromotive force. This force is developed because of changes in the electric field.

## How to Calculate Electromotive Force

The problems of calculating emf in a circuit is mentioned in this section.

### Solved Examples

Question 1: A cell of 3V emf and 0.5$\Omega$ internal resistance is connected to a bulb which has a resistance of 2$\Omega$. Calculate the terminal voltage and current?

Solution:

The given parameters are,
E = 3V, r = 0.5$\Omega$ and R = 2$\Omega$

Equation of the current is,
I = $\frac{E}{R+r}$
I = $\frac{3}{2+0.5}$
I = 1.2A

The terminal voltage is given by,
E - Ir = 3 - (1.2 × 0.5)
=3 - 0.6 = 2.4V

Question 2: A cell of 1.5V emf and 0.2$\Omega$ internal resistance is connected to a bulb which has a resistance of 3$\Omega$. Calculate the terminal voltage and current?

Solution:

The given parameters are,
E = 1.5V, r = 0.2$\Omega$ and R = 3$\Omega$

Equation of the current is,
I = $\frac{E}{R+r}$
I = $\frac{1.5}{3+0.2}$
I = 0.4687A

The terminal voltage is given by,
E - Ir = 1.5 - (0.4687 × 0.2)
=1.5 - 0.0937 = 1.406V