Ohm's Law and Its Limitations: Explanation & Applications

Shivam Yadav logo

Shivam Yadav

| Updated On - Dec 30, 2025

Ohm's Law is named after the German physicist Georg Simon Ohm, who formulated it in 1827. This law is essential for understanding how electricity behaves in various materials and devices. However, like many scientific principles, it has its limitations and doesn’t apply universally.

  • It states that the “​electric current passing through a conductor is inversely proportional to the resistance and directly proportional to the voltage”.
  • Ohm’s law describes the relationship between electric conductors, resistance, and voltage.
  • There are some limitations of Ohm’s law that include its non-applicability in the case of unilateral networks, non-linear objects, non-metallic conductors, and complicated circuits.

Related Links:

Ohm's Law and Its Limitations

What is Ohm’s Law?

Ohm’s law can be defined when the potential difference applied at the ends of the conductor is ‘V,’ and the current moving through it is ‘I, ’ and other conditions, such as temperature, remain constant, such that the potential difference is directly proportional to the current. Mathematically, we have:

V ∝ I

Where:

  • V is the voltage in volts (V),

  • I is the current in amperes (A),

  • R is the resistance in ohms (Ω).

  • This relationship assumes that the temperature and other physical conditions remain constant. 

  • To verify Ohm’s Law experimentally, a simple circuit can be set up with a power source, a resistor, and measuring instruments. 

  • In this setup, by varying the voltage and measuring the current, you can observe the linear relationship.

Graphical Representation of Ohm’s Law

The graphical representation of Ohm’s Law is the straight-line graph obtained when plotting voltage (V) on the y-axis against current (I) on the x-axis for an ohmic resistor. The slope of this line represents the resistance (R).

Ohm’s Law Graphical Representation

Graphical Representation of Ohm’s Law

Applications of Ohm’s Law

Ohm's Law finds widespread use in everyday technology and engineering. Here are some key applications:

Circuit Design and Analysis

Engineers use Ohm’s Law to calculate the required resistance for limiting current in circuits, such as in LED drivers or power supplies. For instance, to prevent an LED from burning out, a series resistor is chosen based on the supply voltage and desired current.

Electrical Appliances

In household appliances like toasters or heaters, Ohm’s Law helps determine power consumption. Power (P) is related by P = V x I, and substituting Ohm’s Law gives P = I2 R or P = V2/R. This is crucial for safety rating and energy efficiency.

Battery and Power Systems

When designing solar panels or battery chargers, Ohm’s Law ensures optimal matching of voltage and resistance for maximum efficiency. It’s also used in automotive electrical systems to troubleshoot issues like faulty wiring.

Electronics and Computing

In microelectronics, Ohm’s Law governs the behavior of resistors in integrated circuits, aiding in signal processing and data transmission. It’s fundamental in understanding transistor biasing, though transistors themselves may not strictly follow the law.

Medical Devices

Devices like defibrillators or ECG machines rely on precise control of current and voltage, guided by Ohm’s Law, to ensure patient safety. These applications highlight how Ohm’s Law underpins modern technology, from smartphones to electric vehicles.

Components of Ohm’s Law

The three main components of Ohm’s Law have been discussed below -

Electric current

The flow of any electrically charged particles, such as electrons, protons, or ions, through a conducting material is known as current.

The SI unit of current is the ampere, which is denoted by the letter ‘A’.

Coulombs per second is used to measure the magnitude of electric current passing through a circuit.

Consequently, an Ampere is equal to one coulomb of charge per second or 6.2 x 1018 electrons per second.

Read More: Difference Between Ammeter and Voltmeter

Voltage

Voltage can be defined as the electrical pressure at two points of an electrical circuit that is applied so that the electric current can move around the same.

Resistance

Resistance, in simple words, can be defined as the obstacles or hindrances faced by the electrons flowing through a circuit.

How to Find Unknown Values of Resistance?

The constant ratio gives the unknown values of resistance. The resistance depends on the length l and the area of cross-section A for a wire of uniform cross-section. It also depends on the temperature of the conductor. At a given temperature, the resistance,

R = ρlA

where ρ denotes the specific resistance or resistivity and is characteristic of the material of the wire. Using the last equation,

V = I × R = IρlA

I/A is the current density and is denoted by j. The SI unit of current density is A/m². So,

E I = j ρ I

This can be written as E = j ρ or j = σE, where σ is 1/ρ is the conductivity.

Limitations of Ohm's law

Ohm’s law is not applicable in the case of non-linear objects. In these components, the current is not proportional to the voltage applied. This is because, for each value of voltage and current, these components have different resistance values. Examples of non-linear components include the thyristor.

Other limitations of Ohm’s law include:

Examples of Limitations of Ohm’s Law

Some of the examples where Ohm’s law fails to apply are:

A diode demonstrates the limitations of Ohm’s law. When the voltage versus current graph for a diode is plotted, it is observed that the relation between current and voltage is not linear. This happens when voltage is marked in a reverse direction so that the magnitude is fixed. The current is therefore produced in the opposite direction with a different magnitude.

Ohm’s law does not apply to a water volt-ammeter as it is a unilateral network.

It is also not necessary that all the conductors obey Ohm’s law. Ohm’s law is not followed by semiconductors such as Germanium and Silicon. Therefore, they are regarded as non-Ohmic conductors.

Frequently Asked Questions on Ohm’s LawQues. Ohm's law provides the relationship between which two components? [1 mark]

  1. Electric Current and Voltage
  2. Electric Current and Potential Difference
  3. Electric Current and Magnetic Field
  4. All of the Above

Ans. The correct answer is “b”: Electric Current and Potential Difference.Ques. Assuming that the resistance of an electric iron is 50 Ω, with a current of 3.2 A flowing through the resistance. Determine the voltage between two points. [2 marks]Ans. As we already know, the formula to find Voltage,⇒V = I × RNow, upon substitution of the values, we get⇒V = 3.2 A × 50 Ω= 160 V

Uses of Ohm’s Law

While there are certain limitations of Ohm’s law, it also has many uses, such as –

  • Ohm’s law is used to calculate the values of current, voltage, or resistance.
  • Using Ohm’s law formula, any one of the values can be derived if the other two are given.
  • Power consumption can also be calculated using Ohm’s law formula.
  • The knob used to control the speed of fans at home is based on the workings of Ohm’s law. By regulating the resistance of the current passing through the fan can be controlled with the help of the regulator. Furthermore, by rotating the circular knob, variable resistance can be achieved. Thus, the amount of power flowing through the fan at any point can be decided after calculating the current and resistance using Ohm’s law.
  • The power flow in household electrical appliances such as heaters, irons, and kettles is computed with the help of Ohm’s Law
  • Ohm's Law can be utilized to approve the static estimations of circuit parts, current levels, voltage supplies, and voltage drops. If a test instrument distinguishes a higher than the ordinary current value, it could imply that resistance has diminished or that voltage has increased, causing a high-voltage circumstance. This could show a stockpile or circuit issue.
  • Ohm’s law is also used in designing and studying DC (direct current) measuring devices such as ammeters and voltmeters.
  • The Ohm’s law formula is used by many devices to calculate information such as the current passing through, the resistance of conductors, and the voltages applied required to build a circuit.

FAQs

Ques. Name a device that we can use to maintain a potential difference between the ends of a conductor. And explain the process by which this device does so. (2 Marks)

Ans. A cell or a battery can be used in order to maintain a potential difference between the ends of a conductor. Even when no current is drawn from it, the chemical reaction within a cell generates the potential difference across the terminals of the cell. When it is connected to a conductor, it produces electric current and, maintain the potential difference across the ends of the conductor.

Ques. A current of 20 A flows through a conductor for two minutes. (i) Find out the amount of charge passed through any area of cross section of the conductor. (ii) If the charge of an electron is 1.6 × 10-19 C, then what is the total number of electrons flowing? (2 Marks)

Ans. Given that: I = 20 A, t = 2 min = 2 × 60 s = 120 s

(i) Amount of charge Q passed through any area of cross-section is given by I = Qt

or, Q = I × t ∴ Q = (20 × 120) A s = 2400 C

(ii) Since, Q = ne

where n is the total number of electrons flowing and e is the charge on one electron

∴ 2400 = n × 1.6 × 10-19

or n = 2400/1.6×10-19 = 7.5 × 1021 = 1.5× 1022

Ques. An electric heater of resistance 4 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater. (3 Marks)

Ans. Rate of heat produced by a device is given by the expression for power as, P= I2R

Where,

Resistance of the electric heater, R= 4 Ω

Current drawn, I = 15 A

P= (15)2 x 4 = 1200 J/s

Therefore, heat is produced by the heater at the rate of 1200 J/s.

Ques. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases? (5 Marks)

Ans. Supply voltage, V= 220 V

Resistance of one coil, R= 24 Ω

(i) Coils are used separately

According to Ohm's law, V= I1R1

Where,

I1 is the current flowing through the coil

I1 = V/R1 = 220/24 = 9.166 A

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω

According to Ohm's law,V = I2R2

Where,

I2 is the current flowing through the series circuit

I2 = V/R2 = 220/48 = 4.58 A

Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel

Total resistance, R3 is given as =

1/(1/24+/1/24)=24/2

=12Ω

As per Ohm's law,

V= I3R3

Where,I3 is the current flowing through the circuit I3 = V/R3 = 220/12 = 18.33 A

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

Ques. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V. (3 Marks)

Ans. Given Charge, Q = 96000C

Time, t= 1hr = 60 x 60= 3600s

Potential difference, V= 50volts

Now we know that H= VIt

So we have to calculate I first

As I= Q/t

∴ I = 96000/3600 = 80/3 A

H=50×(80/3)×60×60

=4.8×106J

Therefore, the heat generated is 4.8 x 106 J.

CBSE CLASS XII Related Questions

  • 1.
    A charged particle is moving in a circular path with velocity \( V \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path change? (2) its time period of revolution change?


      • 2.
        The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

          • \( \frac{\alpha}{r} \)
          • \( \frac{\alpha r^2}{2} \)
          • \( \frac{\alpha}{2r^2} \)
          • \( -\frac{\alpha}{r} \)

        • 3.
          A charge \( -6 \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \mu C \) is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.
          work done on the charge


            • 4.
              A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the \( +x \)-direction. A magnetic field \( \vec{B} = (0.4 \, \text{mT} \hat{j}) + (0.6 \, \text{mT} \hat{k}) \) is switched on. The force acting on the segment is:

                • \( (2 \hat{i} + 3 \hat{k}) \, \text{mN} \)
                • \( (-3 \hat{j} + 2 \hat{k}) \, \mu\text{N} \)
                • \( (6 \hat{j} + 4 \hat{k}) \, \text{mN} \)
                • \( (-4 \hat{j} + 6 \hat{k}) \, \mu\text{N} \)

              • 5.
                A ray of light is incident on a refracting face AB of a prism ABC at an angle of \( 45^\circ \). The ray emerges from face AC and the angle of deviation is \( 15^\circ \). The angle of prism is \( 30^\circ \). Show that the emergent ray is normal to the face AC from which it emerges out. Find the refraction index of the material of the prism.


                  • 6.

                    Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \). 

                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show