Content Strategy Manager | Updated On - Oct 28, 2024
Ohm’s Law of Current Electricity is named after the scientist “Ohm”. 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.
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Key Terms: Current Electricity, Electric Charge, Ohm’s law, Resistance, Ohm’s law Limitations, Voltage, Circuits, Potential Difference.
What is Ohm’s Law?
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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 |
Ohms Law Explained
Ohms law video
It can also be represented as:
| \(I =\frac{V}{R}\) or \(V=I\times R\) |
Limitations of Ohm's law
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The limitations of Ohm’s law include:
- Ohm’s law cannot be applied to unilateral networks. Unilateral networks allow the current to flow only in one direction. Examples of such networks include diodes, transistors, etc.
- Additionally, 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.
- Ohm’s Law will not work in the case of non-metallic conductors.
- Calculation using Ohm’s law can be difficult in the case of complicated circuits.
Examples of Limitations of Ohm’s Law
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Some of the examples where the 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 is not applicable 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]
Ans. The correct answer is “b”: Electric Current and Potential Difference. Ques. Assuming that the resistance of an electric iron is 50 Ω with current 3.2 A flowing via resistance. Determine the voltage between two points. [2 marks] Ans. As we already know the formula to find Voltage, |
Components of Ohm’s Law
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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 S.I unit of current is 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.
- In other words, the voltage can also be known as the difference in charge between two points of a circuit.
- The S.I unit of voltage is Volt which is represented by the letter ‘V’.
Resistance
Resistance in simple words can be defined as the obstacles or hindrances faced by the electrons flowing through a circuit.
- The SI unit of resistance is ohms written as Ω.
- However, the degree of resistance offered by different objects can vary based on the material it is made of as well as the dimensions.
- For example, metals have low resistivity whereas materials like plastics or rubbers offer high resistance to electric current.
- Accordingly, based on their resistance power, objects are classified into three types - insulators, conductors, and semiconductors.
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 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 = σ E, where σ is 1/ρ is conductivity.
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Uses of Ohm’s Law
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While there are certain limitations of Ohm’s law, it has also 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 are 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. In the event that, a test instrument distinguishes a higher than the ordinary current value, it could imply that resistance has diminished or that voltage has expanded, 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 voltages applied required to build a circuit.
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Current Electricity Detailed Video Explanation:
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Things to Remember
- Ohm’s law provides a relation between voltage, current, and resistance.
- The limitations of Ohm’s law include non-applicability in unilateral networks, non-linear objects, non-metallic conductors, and complicated circuits.
- However, it is not necessary that all devices follow ohm’s law. There are many non-ohmic devices.
- Ohm’s law states that the current is directly proportional to voltage.
- Ohm’s law is used to calculate the values of current, voltage, or resistance.
Previous Year Questions
- An example of a non ohmic device is … [KCET 1996]
- A galvanometer of resistance G is converted … [JEE Mains 2020]
- A wire of resistance 12 ohm is in the form of a circle … [JIPMER 1999]
- A galvanometer of 25 ohms and having full scale deflection for a current of 10 mA … [JIPMER 1996]
- A galvanometer has a coil of resistance 100 ohm … [NEET 2010]
- A galvanometer of 50 ohm resistance has 25 divisions … [NEET 2004]
- A parallel plate capacitor has a uniform electric field E … [NEET 2012]
- A resistance wire connected in the left gap of a metre bridge … [NEET 2020]
- A short electric dipole has a dipole moment of … [NEET 2020]
- The potential difference that must be applied to stop the fastest photo electrons … [NEET 2010]
- The potential difference (VA−VB) … [NEET 2016]
- The potential differences across the resistance, capacitance and inductance are … [NEET 2016]
- A and B are two conductors carrying a current I … [KCET 2002]
- The current passing through the ideal ammeter … [KCET 2007]
- The potential difference between A and B … [KCET 2008]
- The amount of charge flowing per second per unit area … [KEAM]
- Find the current in the 8W resistance …
- To send 10% of main current through a moving coil galvanometer … [KCET 2003]
- Two parallel wires carry electric current in same direction … [KCET 1996]
Sample Questions
Ques. Find the current if Volt = 6 V and Resistance is 2 Ω. (2 Marks)
Ans. According to Ohm’s law,
V = R*I
So, 6 = 2(I)
or , I = 6/2 = 3 A
Ques. State Ohm’s Law. (2 Marks)
Ans. Ohm’s law states that if 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 remains constant, then at that point, the potential difference is directly proportional to the current.
Ques: Find the Voltage if Current = 8 A and Resistance is 4 Ω. (2 Marks)
Ans. According to Ohm’s law,
V = R*I
So, V = 4 x 8
or , V = 32
Ques. What are some of the uses of Ohm’s Law? (2 Marks)
Ans. Some of the uses of Ohm’s Law are:
- By using Ohm’s law formula we can calculate the values of current, voltage or resistance.
- Power consumption can also be calculated using Ohm’s law formula.
- The Ohm’s law formula is used by many electrical appliances like heater, iron, oven, etc to calculate voltage, electric current and resistance.
- Ohm's Law can be utilized to approve the static estimations of circuit parts, current levels, voltage supplies, and voltage drops.
Ques. Find the Resistance if Current = 8 A and Voltage is 10 V. (2 Marks)
Ans. According to Ohm’s law,
V = R*I
So, 10 = R x 8
or , R = 10/8
R = 1.25 Ω
Ques. In a given circuit, the current flowing through is 0.01 A when the applied voltage is 5 V. If the voltage applied is increased to 7.5 V, calculate the current flowing through the same. (3 Marks)
Ans. Ohm’s law states that V = RxI
Then, 5 = R x 0.01
We need to find R
R = 5 x 0.01 = 500 Ω
Now that we know the resistance value, we can calculate I when volt is 7.5.
I = 7.5/500 = 0.015 A
Ques. Find the Current if Resistance = 8 Ω and Voltage is 16 V . (2 Marks)
Ans. According to Ohm’s law,
V = R*I
So, 16 = 8 x I
or , I = 16/8
I = 2 A
Ques. Consider an electric circuit with resistors R1 and R2 in series having resistance value of 5 Ω and 10 Ω respectively. If the voltage applied across resistance R1 is equal to 4 V, then calculate the current flowing through voltage and resistance R2. (2 Marks)
Ans. Here we should first calculate the current passing through Resistor R1 using Ohm’s law -
V = R*I
4 = 5*I
Or, I = 4/5 = 0.8 A
Since it is mentioned in the problem above that the two resistors are in series, the current passing through R2 will also be the same as R1.
Ques. How much resistance is required to limit the current from a 12 V battery to 0.04 A? (3 Marks)
Ans. Given a 12 V battery, the task is to find the resistance such that the current will be limited to 0.04 A.
- Voltage V=12 V
- Current I=0.04 A
From Ohm's law,
V=IR
We solve for the resistance and substitute the given values,
R=V/I
= 12 V/0.04 A
R = 300 Ω
Ques. What is the reciprocal of resistivity of a material called? Give its unit. (4 Marks)
Ans. Conductivity = \(\frac{1}{resistivity}\)
SI unit of conductivity = \(\frac{1}{ohmmeter}\)
= ohm-1 m-1.
Ques. Resistance of an electric iron 50 Ω.4.2A Current flows through the resistance. Find the voltage between two points.
Ans. Here, Resistance, R = 50 Ω.
Current, I =4.2 A
Voltage, V =?
From Ohm’s law,
V = IR
= 4.2 × 50
= 210V
Ques. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? (4 Marks)
Ans. According to Ohm’s law
V = IR
⇒ I=V/R ...(1)
Now Potential difference is decreased to half
∴ New potential difference Vʹ=V/2
Resistance remains constant
So the new current Iʹ = Vʹ/R
= (V/2)/R
= (1/2) (V/R)
= (1/2) I = I/2
Thus, the amount of current flowing through the electrical component is reduced by half.
Ques. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, a 12 Ω resistor, and a plug key, all connected in series. (3 Marks)
Ans. Three cells of potential 2 V, each connected in series therefore the potential difference of the battery will be 2 V + 2 V + 2 V = 6V.
The circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V and a closed plug key means the current is flowing in the circuit.
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
\(H = 50 \times \frac{80}{3} \times 60 \times 60 = 4.8 \times 10^6 J\)
Therefore, the heat generated is 4.8 x 106 J.
Ques. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. (3 Marks)
Ans. Resistance (R) of a resistor is given by Ohm's law as, V= IR
R= V/I
Where, potential difference, V= 12 V
Current in the circuit, I= 2.5 mA = 2.5 x 10-3 A
\(R = \frac{12}{2.5 \times 10^{-3}} = 4.8 \times 10^3 \Omega = 4.8 k\Omega\)
Therefore, the resistance of the resistor is 4.8 kΩ
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 = \(\frac{1}{\frac{1}{24}+\frac{1}{24}} = \frac{24}{2} = 12\Omega\)
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. An electric heater of resistance 8 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= 8 Ω
Current drawn, I = 15 A
P= (15)2 x 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.
Ques. Compare the resistances of two wires of the same material. Their length is in the ratio of 2:3 and their diameters are in the ratio of 1:2. (3 Marks)
Ans. R1 = \(\frac{pl_1}{A_2} = \frac{p(2x)}{\pi (y/2)^2} = \frac{8 px}{\pi y^2}\)
R2 =
⇒ R1 : R2 = 8 : 3
Ques. A current of 10 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 = 10 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 = (10 × 120) A s = 1200 C
(ii) Since, Q = ne
where n is the total number of electrons flowing and e is the charge on one electron
∴ 1200 = n × 1.6 × 10-19
or n = 12001.6×10−19 = 7.5 × 1021
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.
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