Electromotive Force often called EMF is the potential difference across the terminal of a cell or a battery when no current is being drawn from it. EMF is a misnomer i.e., it is actually a Potential Difference rather than a force but at the same time, EMF also differs from the Potential Difference in some manners. In this article, we will learn about EMF i.e., Electromotive Force, its formula and how does it differs from the potential difference in detail.
EMF Definition
EMF i.e., Electromotive Force is defined as the potential difference across the terminal of a cell or a battery when no current is being drawn from it. We can also say that it is the maximum voltage across the terminals of the power source in an open circuit. Here, the EMF is a function of the internal resistance of the battery. EMF is the reason behind the flow of current in a circuit from the terminal of higher potential to the terminal of lower potential. The terminal of higher potential is positive and that of lower potential is the negative terminal.
EMF Equation
Electromotive Force is basically a potential difference hence physically it is also equal to the work done per unit charge. We know that work done is physically equal to Energy. Hence, Electromotive Force is equal to the Energy stored in the cell per unit of charge. Hence, Electromotive Force is given as
ε = E/Q
Where,
- ε is the Electromotive Force,
- E is the Energy Stored, and
- Q is the Charge.
We also know that EMF is a potential difference, hence we can also write EMF using Ohm's Law. Hence by Ohm's Law
ε = IR
Where,
- I is the current flowing in the circuit, and
- R is the total resistance of the circuit
But we know that total resistance R is the sum of internal resistance 'r' and resistance offered by the circuit R1. Hence, we can rewrite the above equation as
ε = I(R1 + r)
⇒ ε = IR1 + Ir
ε = V + Ir
Thus, we can say that EMF is the sum of total voltage across the circuit and the product of the current and internal resistance of the cell.
Formula of Electromotive Force
The formula of EMF is given in two ways. If Energy and Charge are given then we calculate EMF by taking the ratio of Energy to the charge i.e., ε = E/Q.
If Voltage, Current and internal resistance are given then the formula of EMF is given as ε = V + Ir
Unit of EMF
Since, in physical terms, EMF is equivalent to Potential Difference, thus its unit will also be the same as that of Potential Difference. Hence, the unit of EMF is Volt(V) which is equal to Joule per Coulomb(J/C).
Dimension of EMF
Dimensionally EMF is equal to Potential Difference. Hence the dimensional formula of EMF is [M1L2T-3I-1]. To calculate the dimensional formula of EMF, divide the Dimensional Formula of Energy by Charge. The dimensional formula of Energy is equivalent to that of work i.e., [ML2T-2] and the dimensional formula of charge is [IT]. Hence, the ratio of [ML2T-2] and [IT] is [M1L2T-3I-1] which is the dimensional formula of Electromotive Force.
Difference Between Potential Difference and EMF
The basic difference between the Potential Difference and EMF is that EMF is the potential difference across the terminals of the battery when no current is drawn from the circuit while Potential Difference is valid when current is being passed through an appliance or a load. EMF gives an idea of the conversion of any form of energy whether chemical, mechanical, or anything else to electrical energy per unit of charge but potential difference gives an idea of work done by electrical energy per unit of charge.
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Sample Problems
Problem 1. Calculate the electromotive force of a circuit with energy and charge of 1200 J and 20 C respectively.
Solution:
We have,
E = 1200
Q = 20
Using the formula we get,
ε = E/Q
⇒ ε = 1200/20
⇒ ε = 60 V
Problem 2. Calculate the charge of a circuit with energy and the terminal voltage of 300 J and 10 V respectively.
Solution:
We have,
E = 1200
ε = 10
Using the formula we get,
ε = E/Q
⇒ Q = E/ε
⇒ Q = 300/10
⇒ Q = 30 C
Problem 3. Calculate the energy of a circuit with charge and the terminal voltage of 25 C and 15 V respectively.
Solution:
We have,
Q = 25
ε = 15
Using the formula we get,
ε = E/Q
⇒ E = εQ
⇒ E = 15 (25)
⇒ E = 375 J
Problem 4. Calculate the electromotive force of a circuit with a potential difference of 3 V, with a current of 0.5 A. The internal resistance of the battery is 1 ohm.
Solution:
We have,
V = 3
I = 0.5
r = 1
Using the formula we get,
ε = V + Ir
⇒ ε = 3 + 0.5 (1)
⇒ ε = 3 + 0.5
⇒ ε = 3.5 V
Problem 5. Calculate the potential difference of a circuit with a terminal voltage of 4 V, with a current of 2 A. The internal resistance of the battery is 0.2 ohm.
Solution:
We have,
ε = 4
I = 2
r = 0.2
Using the formula we get,
ε = V + Ir
⇒ V = ε - Ir
⇒ V = 4 - 2 (0.2)
⇒ V = 4 - 0.4
⇒ V = 3.6 V
Problem 6. Calculate the electromotive force of a circuit with a current of 1.5 A and load resistance of 2 ohm. The internal resistance of the battery is 1 ohm.
Solution:
We have,
I = 1.5
R = 2
r = 1
Using the formula we get,
ε = IR + Ir
⇒ ε = 1.5 (2) + 1.5 (1)
⇒ ε = 3 + 1.5
⇒ ε = 4.5 V
Problem 7. Calculate the load resistance of a circuit with a terminal voltage of 7 V and a current of 3 A. The internal resistance of the battery is 2 ohm.
Solution:
We have,
I = 3
ε = 7
r = 2
Using the formula we get,
ε = IR + Ir
⇒ IR = ε - Ir
⇒ 3R = 7 - 3 (2)
⇒ 3R = 7 - 6
⇒ R = 1/3
⇒ R = 0.33 ohm