The flow of electric charges is known as electric current. In earlier times, it was considered that correct flow was due to the movement of positive charges from the positive terminal to the negative terminal. But after the discovery of electrons, the conventional wisdom was proven wrong as the positive charge is nothing but the deficiency of the electron. Now, scientists know that correct itself is generated due to the movement of negative charges (electrons), electron flow from the negative terminal of the system to the positive terminal, which is known as Electron flow, and unlike electrons, current flows in the opposite direction i.e., from the positive terminal to negative terminal, which is also known as Conventional Current flow.
The best way to imagine current flowing in a circuit is by imagining the flow of liquid inside a pipe from a higher point to a lower one, the liquid represents current, and the difference in the heights of the extreme end of the pipe represents the difference in potential causing the current to flow if the pipe is twisted at some point the flow of the liquid will slow down, that represents the resistance offered by the conductor that reduces the current.
Electric Current Formula
The electric current can be represented as the rate of flow of electric charge (q) which mathematically can be represented as follows:
As Electric Charge and Time come under the Fundamental quantities and their units are respectively Coulomb and second. So by the definition of Electric Current, its unit is Coulomb/second.
The SI Unit of electric Current is Ampere (A).
In CGS system, unit of electric current is biot or sometimes called an abaampere.
What is EMF?
Normally, in a conductor, Electrons are present, and they are not stationary, they keep on moving in random directions, and due to their randomness, the overall displacement of all electrons becomes zero and hence, no current is produced. In order to produce current, some external force is required to align electrons in one direction and make them move in that one direction, this external force is known as Electromotive Force and is also famous as EMF. It is nothing but the voltage applied to produce current.
Types of Electric Current
There are two types of electric current, which are as follows:
Alternating Current
Electric current with its direction and values keep changing is known as Alternating Current. The values of AC in one direction increase from 0 to peak value then falls down to 0 again, then in opposite direction increase from 0 to peak value then come back to 0.
As electric current goes from 0 to peak in one direction and the same in opposite direction, it behaves like a wave more specifically a sine wave and it has some frequency. For example, in India, the grid provides 60 Heartz 220 Volts AC but in America, their grid provides 120 Volts 50 hearts AC.
Direct Current
Electric current with the same direction always is known as Direct Current. As the direction of DC remains the same, so its frequency is 0.
Properties of Electric Current
Current is due to the Flow of electrons in the circuit.
Electric current can be categorized as AC and DC in nature, where DC is the direct current that flows in only one direction, DC is used in low-voltage applications, aircraft applications, etc. AC is known as alternating current, and it flows in both directions alternatively, AC is the current that comes in our houses and the appliances work on AC.
The electric current in a circuit can be controlled by introducing resistance to the circuit.
The unit of Electric current is Amperes (A). 1 Ampere can be defined as the flow of 1 coulomb of charge in 1 second.
Electric current flows from Higher potential to Lower Potential in a circuit (from positive terminal to negative terminal) also known as the conventional current flow direction.
Ohm’s Law
The German physicist Georg Simon Ohm stated that the current flowing in a wire is directly proportional to the voltage drop across it. According to Ohm’s Law, the current flowing through a wire is directly proportional to the voltage applied at the ends of the wire provided that the temperature and conductivity remain the same.
V∝I
Upon removing the proportionality sign, a proportionality constant is introduced known as Resistance.
Where,
V is Voltage at the ends of conductor
R is Resistance offered by the conductor
I is Current through the wire.
Ohm’s law can be illustrated using the following illustration,
Effect of Electric Current
There are different effects that can be noticed due to the flow of electric current in a wire, for example, when current passes through a resistor, the resistor has a property of resisting which does not let the whole current pass but since energy can neither be created nor destroyed, it is converted in heat energy and is released in the form of heat, this effect is called as the heating effect of current. Similarly, we have the magnetic and chemical effects of electric current.
Chemical effect of Electric Current
When Electric current is passed through a which is conducting in nature, the solution breaks in its respective ions, and effects are seen visibly. The major effects that are prominent,
The color of the solution may change.
The deposition of metal at the electrodes may be seen.
There can be the formation of gas bubbles at the electrodes.
Magnetic effect of Electric current
Electric current is nothing but the motion of electrons, and it is known, when charges are stationary, they create Electric Field but when charges are in motion, they create a Magnetic field. When current is passed through a wire and a metallic sheet is placed there with a needle, the needle will be deflected due to the presence of a magnetic field which is produced by the electric current. One of the biggest applications of the Magnetic effect of electric current is Electromagnets, they are formed with the help of passing current.
Heating effect of Electric Current
When current flows in a conductor, heat energy is produced and released from the conductor and the amount of effect depends upon the resistance offered by the conductor. If the conductor has to offer high resistance, it simply means that it shall not allow most of the current to flow but due to the conservation of energy (energy can neither be created nor destroyed), current that could not pass is converted into heat and the phenomenon is known as the Heating effect of current. The formula for heat energy is given by,
Where, H is Heat energy released
I is current flowing in the conductor
R is Resistance offered by the conductor
T is Time for which the current was flowing in the conductor.
Applications of the heating effect of current involve Electric Irons, Electric Heaters, filament lamps, Electric kettles, etc.
Solved Examples of Electric Current
Question 1: In a conductor, 10 Coulombs of charge flow for 5 seconds, determine the current produced.
Solution:
The current in a circuit is given by,
I = q/t
⇒ I = 10/5 Amperes
⇒ I = 2 Amperes
Therefore, 2 amperes of current flows in the circuit.
Question 2: Which material is a better conductor of electricity — Iron or Silicon?
Answer:
Iron is a better conductor of electricity while silicon is a semiconductor.
Question 3. What is the best conductor of electricity?
Answer:
Silver is known to be the best conductor of electricity.
Question 4:In the circuit given below, Find the current flowing through the circuit.
Solution:
In the figure provided, it is clear that there are two resistances, and they are in series. When two or more than two resistances are attached in series in a circuit, the overall resistance becomes equal to the sum of individual resistances present in the circuit.
R = R1+ R2
⇒ R = 2+ 2
⇒ R = 4 ohms
From Ohms Law, we know
V = IR
⇒ I = V/R
⇒ I= 20/4
⇒ I= 5 Amperes
Question 5: Why Silver is not used more often as a conductor in daily usage?
Answer:
Silver is known to be the best conductor of electricity but is still not preferred to be used in daily usage as a conductor because of its unavailability and the fact that silver is very expensive. Instead, conductors like copper which are easily available are used.
Question 6: What is the Heat energy produced when 2 amperes of current is flowing in a circuit for 5 seconds having an overall resistance in the circuit of 4 ohms?
Solution:
The Heat energy produced is given by,
H= I2RT
⇒ H= (2)2×4 × 5
⇒ H= 16 × 5
⇒ H= 80 Joules
Therefore, 80 Joules is produced in the circuit.
FAQs on Electric Current
Question 1: What is Electric Charge?
Answer:
Electric charge is the fundamental property of matter that exists due to the presence or absence of an electron in an atom. It is a physical property that tells us how strangely an object will behave when placed in an electromagnetic field.
Question 2: What is Electric Current?
Answer:
Rate of flow of electric charge in a closed is known as Electric Current.
Question 3: What is the si unit of electric current?
Answer:
SI unit of electric current is Ampere.
Question 4: Define ampere.
Answer:
In a conductor if 1 Coulomb of charge passes in 1 second of time, then we say 1 Ampere current is flowing through that conductor.
Question 5: An instrument that detects electric current is known as________.
Answer:
Instrument for measuring electric current is known as Ammeter.
Electric Potential
Electric potential is the work done per unit charge in order to bring that charge from infinity to a point in the electrostatic field against the field force. Electric Potential is also referred to as Voltage drop. The SI unit for Electric Potential or Electric Potential difference is Voltage or Volts. Electric potential is a scalar quantity.
In the above figure, +Q is the charge creating an electric field, and the task is to bring a unit charge (+q) from infinity (anywhere outside the electric field) to a point inside the electric field against the field. Since the electrostatic force produced by the electric field will be against the unit charge, some work is required by the unit charge in order to move from either infinity to somewhere in the field or from one point to another in the field.
If the work is done to move the charge from infinity to point X, it will be called an Electric potential at X (Vx).
If the work is done to move the charge from infinity to point Y, It will be called an Electric potential at Y (Vy).
If the work is done to move the charge from X to Y, it will be called the potential difference between x and y (Vxy).
Electric Potential Formula
Electric Potential/Voltage = Work Done/Unit Charge
Therefore, the SI unit for Electric Potential is Volts or Voltage.
1 Volt = 1 Joule/1 Coulomb
1 Volt can be defined as 1 joule of work done in order to move 1 coulomb of charge
Electric Potential Difference
Electric potential difference is also known as voltage. The electric potential difference is the work done per unit charge to move a unit charge from one point to another in an electric field. Electric potential difference is usually referred to as a Voltage difference. Imagine a ball sitting at some height, will there be some energy in the ball? Yes, the energy is called Potential energy, and if the ball is dropped from a point A to B height, the ball will always fall from higher gravitational potential to lower, then there will be a difference in both energies. The electrical potential difference is analogical to this concept. The energy possessed by Electric charges is known as electrical energy. A charge with higher potential will have more potential energy, and a charge with lesser potential will have less potential energy. The current always moves from higher potential to lower potential. The difference in these energies per unit charge is known as the electric potential difference. In order to create electricity and the flow of current, a potential difference is always required, which is maintained by a battery or a cell.
The longer side represents the Higher potential (+ve terminal), and the shorter side represents the Lower potential (-ve terminal). The Electric potential difference is measured by a Voltmeter which is applied parallel to the Instrument whose Voltage is to be measured.
Electric Potential Difference Formula
Vxy= Vx – Vy = [Wx – Wy]/q
SI Unit of Electrical Potential Difference
The SI Unit of Electrical potential difference is the same as the electric potential, i.e, Voltage or Volts.
Note: Why a unit charge is taken to explain the concept of Electric Potential?
A small unit charge will have very small electric field which will neither affect not distort the field produced by the bigger charge and hence, the concept can be explained easily.
Difference Between Electric Potential and Electric Potential Difference
Electric Potential
Potential Difference
Electric potential is the work done per unit charge to bring the charge from infinity to a point in an electric field.
Potential difference is the difference between the potentials between two points in the electric field.
Electric potential is defined at a point.
Potential difference is defined between two points in electric field.
The unit of electric potential is volts.
The unit of potential difference is volts.
Electric potential is a scalar quantity.
Potential difference is also a scalar quantity.
Electric Potential is taken as zero at infinity.
There is no as such concept for potential difference.
Ohm’s Law
According to Ohm’s Law, The current flowing through a conductor is directly proportional to the voltage, provided the conductivity and temperature remain constant. While removing the proportionality, a new constant is introduced known as the Resistance.
V ∝ I
V = IR
Where,
V = Voltage across the conductor
I = Current flowing in the wire
R = Resistance offered by the conductor in ohms
Ohm’s Law Graph
Ohm’s Law holds good when physical conditions like temperature and others are constant. This is because of the fact that the current flowing through the circuit varies by changing the temperature. Therefore, in such cases when physical factors like temperature come into play, Ohm’s law violates. For example, in the case of a Light bulb, where temperature increases when the current flowing through it rises. Here, Ohm’s Law doesn’t follow.
The graph for an ohmic circuit is discussed in the image below,
Ohms Law Graph
Ohm’s Law Unit
There are three physical quantities that are associated with the Ohms Law that include,
Current
Voltage
Resistance
The table added below shows the various symbol and their unit used.
Physical Quantity
Unit of Measurement
Unit Abbreviation
Current(C)
Ampere
A
Voltage(V)
Volt
V
Resistance(R)
Ohm
Ω
Ohm’s Law Equations
Ohm’s law provides three equations which are:
V = I × R
I = V / R
R = V / I
Where,
V is the voltage,
I is the current, and
R is the resistance.
Relationship between Voltage, Current, and Resistance
The relation between voltage, current, and resistance can easily be studied using the formula,
V = IR
Where,
V is the voltage,
I is the resistance, and
R is the resistance.
We can study this formula with the help of the table discussed below,
Voltage
Current
Resistance
2 V
1/2 A
4 Ω
4 V
1 A
4 Ω
8 V
2 A
4 Ω
Ohm’s Law Triangle
Ohm’s Law Triangle is a visual representation for understanding and learning the Ohm’s Law relation between voltage, current, and resistance. This tool help helps engineers to remember the order of the relationship between the three main aspects: current (I), voltage (V), and resistance (R).
Ohms Law Triangle
Vector Form of Ohm’s Law
The relation between current and voltage is established by, Ohm’s law, and its vector form is,
Where,
is Current Density vector,
is Electric Field vector, and
σ is conductivity of material.
Resistivity
The hindrance faced by the electrons while moving in any material is called the resistivity of the material.
Let a resistor of a length of ‘l’ and the cross-sectional area of ‘A’ has a resistance be R. Then we know,
Resistance is directly proportional to the length of the resistor, i.e. R ∝ l, . . .(1)
Resistance is inversely proportional to the cross-section area of the resistor, i.e. R ∝ 1/A . . .(2)
combining eq. (1) and eq.(2)
R = ρl / A
Where ρ is the proportionality constant called coefficient of resistance or resistivity.
Now if L = 1m and A = 1 m2, in the above formula we get,
R = ρ
This means for a resistor of length 1 m and cross-section area 1 m2 the resistance is called the resistivity of the material.
Experimental Verification of Ohm’s Law
Verification of Ohm’s Law is achieved by performing the following experiment.
Apparatus Required
The apparatus required for performing the experiment for the Verification of Ohm’s Law is,
Resistor
Ammeter
Voltmeter
Battery
Plug Key
Rheostat
Circuit Diagram
The circuit diagram for the Experimental Verification of Ohm’s Law is given in the diagram below,
Circuit Diagram of Ohms Law
Procedure
The procedure for experimental verification of Ohm’s Law is mentioned below:
The key K is closed initially and the rheostat is adjusted such that the reading in ammeter A and voltmeter V is minimum.
The current is then increased in the circuit by adjusting the rheostat, and the current at various values of the rheostat and their respective voltage is recorded.
Now for different values of voltage(V) and current(I) and then calculate the ratio of V/I.
After calculating all the ratios of V/I for different values of voltage and current, we notice that the value is almost constant.
Now plotting a graph of the current against the potential difference we get a straight line. This shows that the current is directly proportional to the potential difference and its slope is the resistance of the wire.
Ohm’s Law Pie Chart
To better understand the relationship between various parameters, we can take all the equations used to find the voltage, current, resistance, and power, and condense them into a simple Ohm’s Law pie chart as shown below:
Ohms Law Pie Chart
Ohm’s Law Matrix Table
Like Ohm’s Law Pie Chart shown above, we can condense the individual Ohm’s Law equations into a simple matrix table as shown below for easy reference when calculating an unknown value.
Ohms Law Matrix Table
Applications of Ohm’s Law
When the other two numbers are known, Ohm’s law may be used to determine the voltage, current, impedance, or resistance of a linear electric circuit.
Main applications of Ohm’s Law:
It also simplifies power calculations.
To keep the desired voltage drop between the electrical components, Ohm’s law is employed.
An electric circuit’s voltage, resistance, or current must be determined.
Ohm’s law is also utilized to redirect current in DC ammeters and other DC shunts.
How to Establish a Current-Voltage Relationship?
The ratio V ⁄ I remains constant for a given resistance while establishing the current-voltage connection, hence a graph of the potential difference (V) and current (I) must be a straight line.
How can we discover the unknown resistance values?
The constant ratio is what determines the unknown resistance values. The resistance of a wire with a uniform cross-section relies on the length (L) and the cross-section area (A). It also relies on the conductor’s temperature.
The resistance, at a given temperature,
R = ρ L ⁄ A
where, ρ is the specific resistance or resistivity and is the wire material’s characteristic.
The wire material’s specific resistance or resistivity is,
ρ = R A ⁄ L
Calculating Electrical Power using Ohm’s Law
We define electric power as the power required by electric charges to do various works. The rate of consuming electric energy is called electric power. The unit of measuring electric power is the watt. Using the Ohm’s law we can easily find the power of the electric circuit. The formula to calculate the electric power is,
P = VI
Where,
P is the power of the circuit,
V is the voltage across the circuit, and
I is the current passing through the circuit.
We know that, using Ohm’s Law,
V = IR
Using the power formula we get,
P = V2/R
P = I2R
Limitations of Ohms Law
Various limitations of the Ohms law are,
The law of Ohm does not apply to unilateral networks. The current can only flow in one direction in unilateral networks. Diodes, transistors, and other electronic components are used in these sorts of networks.
Non-linear components are also exempt from Ohm’s law. Non-linear components have a current that is not proportional to the applied voltage, which implies that the resistance value of those elements varies depending on the voltage and current. The thyristor is an example of a non-linear element.
Analogies of Ohm’s Law
There are various analogies given in the past to explain Ohm’s Law, some of the most common analogies are:
Water Pipe Analogy
Temperature Analogy
Let’s discuss these analogies in detail.
Water Pipe Analogy for Ohm’s Law
We know that the current passing through any circuit depends on the voltage applied and the resistance of the circuit. But we can see the current flowing through the circuit, to understand it better we use the Water Pipe Analogy in which the flowing water represents the current and we can understand Ohm’s law using this concept.
Water flowing through the pipes is similar to the current flowing through the electric circuit. We know that in an electric circuit, Voltage is required to move the current in the circuit in the same way Pressure in the water pipe system allows the water to flow easily in the system.
If the pressure is increased more water flows through the pipe which resembles Ohm’s law that state if the voltage is increased, more current flows through the electric Circuit.
Temperature Analogy
Similarly, a temperature circuit can also be compared to an ohmic conductor. Here, temperature gradient works similarly to voltage, and heat flow works similarly to current.
Example 1: Find the resistance of an electrical circuit with a voltage supply of 15 V and a current of 3 mA.
Solution:
Given:
V = 15 V,
I = 3 mA = 0.003 A
The resistance of an electrical circuit is given as:
⇒ R = V / I
⇒ R = 15 V / 0.003 A ⇒ R = 5000 Ω ⇒ R = 5 kΩ
Hence, the resistance of an electrical circuit is 5 kΩ.
Example 2: If the resistance of an electric iron is 10 Ω and a current of 6 A flows through the resistance. Find the voltage between two points.
Solution:
Given:
I = 6 A, R = 10 Ω
The formula to calculate the voltage is given as:
V = I × R
⇒ V = 6 A × 10 Ω ⇒ V = 60 V
Hence, the voltage between two points is 60 V.
Example 3: Find the current passing through the conductor drawing 20 volts when the power drawn by it is 60 watts.
Solution:
According to Ohm’s P = VI
Given P = 60 watt, V = 20 volt
⇒ I = P/V ⇒ I = 60/20 ⇒ I = 3 A
Hence, the current flowing through the conductor is 3 A
Example 4: A battery of 6 V is connected to the bulb of resistance 4 Ω. Find the current passing through the bulb and the circuit’s power.
Solution:
Given, V = 6 V R = 4 Ω
We know that,
V = IR (Ohms Law)
⇒ 6 = 4R
⇒ I = 6 ÷ 4 = 1.5 A
⇒ I = 1.5 A
Thus, the current flowing through the bulb is 1.5 A
For the Power of the circuit
P = VI
⇒ P = (6)(1.5)
⇒ P = 9 watt
Thus, the power of the circuit is 9 watts.
FAQs on Ohm’s Law
Q1: What is Ohm’s Law?
Answer:
According to Ohm’s Law the current passing through the conductor is directly proportional to the potential difference across the end on the conductor, if temperature and the other physical conditions doesn’t changes.
Q2: Who discovered Ohm’s Law?
Answer:
The German physicist Georg Simon Ohm was the first to explain Ohm’s Law. He stated that the current passing through the conductor is directly proportional to the voltage applied.
Q3: Is Ohm’s Law universally applicable?
Answer:
No Ohm’s law is not a universal law as it is not applicable to all electric circuits.
The circuits which obey Ohm’s Law are called Ohmic Circuit
The circuits which do not obey Ohm’s Law are called Non-Ohmic Circuit
Q4: When was Ohm’s Law Discovered?
Answer:
Ohm’s law was first stated by Georg Simon Ohm in his book Die Galvanische Kette, Mathematisch Bearbeitet in the year 1827.
Q5: What is the unit of Resistance?
Answer:
The SI unit of resistance is Ohm. It is denoted by Ω.
Q6: What is the Dimensional Formula for Resistance?
Answer:
Dimensional formula for resistance is [M1L2T-3I-2]
Q7: Why doesn’t Ohm’s Law apply to semiconductors?
Answer:
Semiconducting devices are nonlinear in nature due to which Ohm’s law does not apply to them. This indicates that the voltage-to-current ratio does not remain constant when voltage varies.
Q8: When does Ohm’s law fail?
Answer:
The behavior of semiconductors and unilateral devices like diodes defines Ohm’s law. If physical factors such as temperature and pressure are not kept constant, Ohm’s law may not provide the intended effects.
Solved Examples on Difference Between Electric Potential and Potential Difference
Question 1: A charge of 10mC is moved from infinity to point A in the Electric field. The work done in this process is 20 Joules. What is the Potential difference?
Solution:
V= W/q
W= 20 joules
q= 10mC
V= 20/10×10-3= 2kV
Therefore, the potential developed is 2000volts.
Question 2: A Charge of 50mC is moved from one point to another (from A to B). The voltage at A is 50kV, and the Voltage at B is 30kV, Find the Work done by the charge.
Solution:
VAB = VA– VB = 50kV – 30kV = 20kV
Work done = V× q
= 20kV × 50mC
= 100 Joules
Question 3: What are the SI units of these following quantities: Energy, Potential Difference, Charge, Resistance
Answer:
The SI Units of the above-mentioned quantities:
Energy⇢ Joules
Potential Difference⇢ Voltage
Charge⇢ Coulomb
Resistance⇢ Ohms
Question 4: Find the current through the circuit when the voltage across the terminal is 30V and the resistance offered by the conductor is 10ohm.
Solution:
According to ohms Law,
V= IR
30V = I × 10ohm
I= 3 Amperes
Therefore, the current through the circuit is 3 amperes.
FAQs on the Difference Between Electric Potential and Potential Difference
Question 1: What is electric potential?
Answer:
Electric potential is defined as the work done to bring a unit positive charge from infinity to that point in the electric field. The SI unit of electric potential is volts.
Question 2: What is the potential difference?
Answer:
Potential difference is the amount of work required to bring a unit charge from one point in an electric field to the other point. The SI unit of potential difference is volts.
Question 3: What is the difference between Electric Potential and Electric Potential difference.
Answer:
Electric Potential is the work done per unit charge in order to bring the charge from infinity to a point in electric field while Electric potential difference is the Potential developed while moving a charge from one point to another in the field itself.
The SI Unit of both electric potential and electric potential difference is Volts/ Voltage.
Question 4: What happens when the battery is in connection with the circuit?
Answer:
When the battery is in connection with the circuit, a potential difference is created at the ends of the conductor. The potential difference is responsible for the formation of an electric field throughout the conductor, and hence the current starts to flow from high potential to low potential.
Question 5: What is the difference between EMF and potential difference?
Answer:
There is difference between EMF and potential difference. EMF is the difference between the potentials of two electrodes, while potential difference is the difference between any two points in the circuit.
Electric Circuit Formula
Various formulas that are used in solving electric circuits are,
Formula
Notations
Electric current
I = Q/t
I is the current flowingQ is the charge flowingt is the time period
Voltage
V = IR
V is the potential difference
Resistance
R = ρ l/A
R is the resistance of the circuitρ is the resistivity of the wirel is the length of the wireA is the cross-sectional area
Power
P = VI or E/t
P is the powerE is the energy gain or losst is the time period
Series Resistance
Req = R1 + R2 + …..+ Rn
Req is the equivalent resistance of the resistors in seriesR1, R2… Rn are the individual resistors added in series
Parallel Resistance
1/Req = 1/R1 + 1/R2 + …..+ 1/Rn
Req is the equivalent resistance of the resistors in parallelR1, R2… Rn are the individual resistors added in parallel
factors on which the resistance of the conductor depends:
1. Length of a Conductor: Consider a Copper wire of length 1 m and connect it between terminals A and B of the circuit. Note the reading of the ammeter. Now take another copper wire of the same area of cross-section but of length 2 m. Connect it between terminals A and B by disconnecting the previous wire. Again, note the reading of the ammeter. It will be found that the reading of the ammeter (i.e., electric current) in the second case is half of the reading of the ammeter in the first case.
Since I = V / R so the resistance of the second wire is double the resistance of the first wire.
It shows that the resistance of a conductor is directly proportional to the length of the conductor.
Thus, more is the length of a conductor, more is its resistance.
2. Area of the cross-section of a Conductor: Now take two copper wires of the same length but of a different area of cross-sections. Let the area of the cross-section of the first wire is more than the area of the cross-section of the second wire. Connect the first wire between terminals A and B in the circuit shown in the figure above. Note the reading of the ammeter. Now disconnect the first wire and connect the second wire between terminals A and B. Again note the reading of the ammeter. It will be found that the reading of the ammeter (i.e., electric current) is more when the first wire (i.e., thick wire) is connected between A and B than the reading of the ammeter when the second wire (i.e., thin wire) is connected between the terminals A and B.
It shows that the resistance of a conductor is inversely proportional to the area of cross-section of the conductor.
Thus, the resistance of a thin wire is more than the resistance of a thick wire.
3. Effect of the Nature of material: Take two identical wires, one of copper and the other of Aluminum. Connect the copper wire between terminals A and B. Note the reading of the ammeter. Now, connect the aluminum wire between terminals A and B. Again note the reading of the ammeter. It is found that the reading of the ammeter when the copper wire is connected in the circuit is more than the reading of the ammeter when the aluminum wire is connected in the circuit.
This implies that the resistance of the copper wire is less than the resistance of the aluminum wire.
Hence, the resistance of a wire or a conductor depends upon the nature of the material of the conductor.
4. Effect of temperature of the conductor: If the temperature of the conductor connected in the circuit increases, its resistance increases.
Conclusion:
Thus, factors on which resistance of a conductor depends are:
(i) its length (l),
(ii) Area of the cross-section (A),
(iii) the nature of its material and
(iv) its temperature.
Resistivity or Specific Resistance
Experimentally, it is found that the resistance of a conductor is:
Directly proportional to its length (l) or R ∝ l.
Inversely proportional to its area of cross-section (A) or R ∝ 1/A.
Let’s combine these two conditions for the resistance as
R ∝ l / A
or
R = ρ × l / A
where ρ is the constant of proportionality and known as the Resistivity or Specific Resistance of the conductor.
Rearrange the above expression for ρ,
ρ = R × A / l
Thus, the Resistivity of a conductor is defined as the resistance of the conductor of unit length and unit area of cross-section.
In other words, the resistivity of the conductor is termed as the resistance offered by 4 cubical conductors of 1 m side to the flow of current across the opposite face of the conductor.
In CGS system the unit of resistivity or specific resistance is ohm-cm (Ω-cm).
In SI unit of resistivity or specific resistance is ohm-meter (Ω-m).
To understand more about the concept and formula of resistivity let’s consider an example as:
Example: A conductor having a resistance of 20 Ohms is drawn out so that its length is increased to twice its original length. Calculate the resistance of the new wire.
Solution:
Given that,
The resistance of the conductor, R is 20 Ω.
If original length of the conductor is l then new length (l’) of the conductor is 2l.
So, if the original cross-sectional area is A and new cross-sectional area is A’, then:
Since, l’ = 2l therefore A’ = A / 2.
In case of the original conductor, original resistance is:
R = ρ × l / A
20 Ω = ρ × l / A …… (1)
Now, in case of the new conductor, new resistance is:
R’ = ρ × l’ / A’
= (ρ × 2l) / (A / 2)
= (4 × ρ × l) / A
Use equation (1) in the above expression and solve to calculate R’.
R’ = 4 × 20 Ω
= 80 Ω
Hence, the resistance of the new wire is 80 Ω.
Sample Problems
Problem 1: Find the resistance of a conductor if the current flowing through it is 0.3 A and the applied potential difference is 0.9 Volts.
Solution:
Given that,
The current flow, I is 0.3 A.
The applied potential difference, V is 0.9 V.
Now, use the formula:
R = V / I
= 0.9 V / 0.3 A
= 3 Ω
Hence, the resistance of a conductor is 3 Ω.
Problem 2: A heater has a resistance of 50 Ω and connected to a 220 Volt supply, calculate the current in the heater.
Solution:
Given that,
The resistance of the heater, R is 50 Ω.
The potential difference, V is 220 V.
Now, use the formula:
V = I × R
Rearrange the above expression for I as,
I = V / R
= 220 V / 50 Ω
= 4.4 A
Hence, the current in the heater is 4.4 A.
Problem 3: The resistivity of iron and mercury is given by 10.0 x 10-8 and 94 x 10-8 Ω.cm respectively. Which one is the better conductor?
Solution:
A material whose resistivity is low is consider as good conductor of electricity.
Hence, Iron is the good conductor than Mercury.
Problem 4: Define 1-Ohm resistance.
Solution:
Resistance of a conductor is said to be 1 ohm if a potential difference of 1 volt across the ends of the conductor makes a current of 1 ampere to pass through it.
Problem 5: Let the resistance of an electrical component remains constant while the potential difference across the ends of the components decreases to half of its former value. What change will occur with current through it?
Solution:
Since, I = V / R, when V’=V / 2
I’ = V / 2R = I / 2
Thus, the current in the components becomes half of its former value.
Resistors in Series Combination:
Two or more resistances are said to be connected in series when they are connected end to end and the same current flows through each of them in turn. In this case, the equivalent or the total resistance equals the sum of the number of individual resistances present in the series combination.
Mathematically, the equivalent resistance of any number of resistances (R1, R2, R3, R4, R5, ……..) connected in series is given as:
Req = R1 + R2 + R3 + R4 + R5 + ……..
Consider a case of three resistances (R1, R2, and R3) connected in series with each other with the corresponding voltage source (V1, V2, and V3) in a circuit shown below:
Resistors in Series
The equivalent current flow through it is I, detected through the ammeter A and key K.
The equivalent potential difference is equal to the sum of the individual potential difference across each resistor, i.e.
Veq = V1 + V2 + V3
The current I through each resistor is the same i.e. I = I1 = I2 = I3
Replace the three resistors connected in series by an equivalent single resistor of resistance Req, such that the potential difference Veq across its terminals, and the current I through the circuit remains the same.
Applying Ohm’s law to the circuit:
Veq = IReq
By applying Ohm’s law to all resistors individually as:
V1 = IR1
V2 = IR2
V3 = IR3
Hence, IR= IR1 + IR2 + IR3
or
Req = R1 + R2 + R3
The current through the circuit will remain the same here.
The equivalent potential difference is the sum of the individual potential difference across each resistor.
As a result, equivalent resistance becomes the sum of individual resistances.
The only disadvantage of a series combination is that, if any resistor in a series combination is disrupted or a failure occurs, the whole circuit is switched off.
The series combination is needed to increase the resistance and to divide high potential differences across many resistances.
Such a combination is used in resistance boxes, decorative lights etc.
Resistors in Parallel Combination:
Two or more resistances are said to be connected in parallel connected when they are connected between two points and each has a different current direction. The current is branched out and recombined as the branches intersect at a common point in such circuits.
Mathematically, the equivalent resistance of any number of resistances (R1, R2, R3, R4, R5, ……..) connected in parallel is given as:
1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 + ……..
Consider a case of three resistances (R1, R2, and R3) connected in parallel with each other with the corresponding voltage source (V1, V2, and V3) in a circuit shown below:
Resistors in Parallel Combination
Here, the current flows through each resistor is different therefore, the equivalent current flown through the circuit is:
Ieq = I1 + I2 + I3
Replace the three resistors connected in parallel by an equivalent single resistor of the parallel combination of resistors be Req.
Now, by applying Ohm’s law to the parallel combination of resistors as:
Ieq = V / Req
On applying Ohm’s law to individual resistors as:
I1 = V / R1
I2 = V / R2
I3 = V / R3
Hence, V / Req = V / R1 + V / R2 + V / R3
or
1 / Req = 1 / R1 + 1 / R2 + 1 / R3
In conclusion, we can say that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
The equivalent current through the circuit is the sum of individual currents through each branch of the circuit.
The potential difference across the two terminal points of the circuit remains the same.
As a result, the reciprocal of equivalent resistance of the circuit is the sum of reciprocal of the individual resistances.
In a parallel circuit, a resistor or some other component can be easily connected or disconnected without disturbing the other components.
In parallel combination, the current flown in the circuit is divided into different branches and hence each component receives the required amount of current.
Here, the equivalent resistance is always lesser than all the individual resistances.
If one of the components fails or shorted, the rest of the components of the circuit works usually.
Sample Problems
Problem 1: How much current will an electric lamp draw from a 220 V source, if the resistance of the lamp is 1000 Ω?
Solution:
Given that,
The source voltage, V is 220 V.
The resistance of the lamp, R is 1000 Ω.
The formula to calculate the current drawn is:
I = V / R
Substitute the given values in the above expression as:
I = 220 V / 1000 Ω
= 0.22 A
Hence, the current drawn through the electric lamp is equal to 0.22 A.
Problem 2: If the resistance of the bulb filament is 200 Ω, How much current will an electric bulb draw from a 220 V source?
Solution:
Given that,
The source voltage, V is 220 V.
The resistance of the bulb, R is 200 Ω.
The formula to calculate the current drawn is:
I = V / R
Substitute the given values in the above expression as:
I = 220 V / 200 Ω
= 1.1 A
Hence, the current drawn through the electric bulb is equal to 1.1 A.
Problem 3: The potential difference between the terminals of an electric bulb is 30 V when it draws a current of 6A from the source. What current will the bulb draw if the potential difference is increased to 120 V?
Given that,
The potential difference across the electric bulb, V is 30 V.
The current drawn, I is 6 A.
The increased potential difference, R’ is 120 V.
According to Ohm’s law, the formula to calculate the resistance is:
R = V / I
= 30 V / 6 A
= 5 Ω
When the potential difference is increased to 120 V the current drawn is:
I’ = V / R’
Substitute the given values in the above expression as:
I’ = 120 V / 5 Ω
= 24 A
Hence, the current drawn through the electric bulb is equal to 24 A.
Problem 4: A wire has a resistance of 4 Ω of some given material with length l and cross-section area of A. How much will be the resistance of another wire with the same material having length l/2 and cross-section area of 2A?
Solution:
Consider the resistance of the first wire as:
R1 = ρl / A
where ρ is the resistivity, l is the length and A is the cross-sectional area of the first wire.
But it is given that the resistance, R1 is 4 Ω.
Therefore,
4 Ω = ρl / A ……(1)
Now, in case of second wire:
The length of the wire, l2 is l/2 and
The cross-sectional area, A2 is 2A.
Therefore, the resistance for second wire becomes:
R2 = ρl2 / A2
= ρ(l/2) / (2A)
= ρl / 4A
Substitute 4 Ω for ρl / A, from equation (1) in the above expression.
R2 = 4 Ω / 4
= 1 Ω
Hence, the resistance of the second wire is 1 Ω.
Problem 5: An electric appliance that is connected to a resistance of 40 Ω, and a conductor of 12 Ω resistance to a battery of 8 V in series. Calculate :
(a) Total resistance of the circuit,
(b) Current through the circuit, and
(c) Potential difference across each resistance.
Solution:
The given problem can be represented diagrammatically as:
(a) Given that,
The resistance of electric lamp, R1 = 40 Ω,
The resistance of the conductor connected in series, R2 = 12 Ω.
Therefore, the total resistance of the circuit is:
R = R1 + R2
Substitute the given values in the above expression.
Req = 40 Ω + 12 Ω
= 52 Ω
Hence, the total resistance in the circuit is 52 Ω.
(b) Again, it is given in the problem that,
The total potential difference across the two terminals of the battery, V is 8 V.
Using Ohm’s law, the current flow in the circuit is:
I = V / Req
I = 8 V / 52 Ω
= 0.15 A
Hence, the current through the circuit is 0.15 A
(c) Again, apply Ohm’s law separately to the electric lamp and conductor, the potential difference across the electric lamp is:
V1 = I × R1
= 40 Ω × 0.15 A
= 6 V
And the potential difference across the conductor is,
V2 = I × R2
= 12 Ω × 0.15 A
= 1.8 V
Problem 6: Consider a circuit consist of three resistors as: R1 = 6 Ω, R2 = 18 Ωand R3 = 36 Ω arranged in parallel combination, through a source battery of 12 V. Calculate:
(a) Current through each resistor,
(b) Total current in the circuit, and
(c) Total circuit resistance.
Solution:
The given problem can be represented diagrammatically as:
The resistances connected in parallel combination are: R1 = 6 Ω, R2 = 18 Ω, and R3 = 36 Ω.
The potential difference across the battery, V is 12 V.
Since, the circuit has resistors connected in parallel, then the potential difference across the terminals of the individual resistor will be same.
Hence, to calculate the current in the resistors, use Ohm’s law as:
I1 = V / R1
= 12 V / 6 Ω
= 2 A
Similarly,
I2 = V / R2
= 12 V / 18 Ω
= 0.66 A
And
I3 = V / R3
= 12 V / 36 Ω
= 0.33 A
Therefore, the total current in the circuit,
I = I1 + I2 + I3
= (2 + 0.66 + 0.33) A
= 2.99 A
Hence, the current through each resistor is 2 A, 0.66 A and 0.33 A and the total current in the circuit is 2.99 A.
The total resistance Req is:
1 / Req = 1 / R1 + 1 / R2 + 1 / R3
= 1/ 6 Ω + 1/ 18 Ω + 1/ 36 Ω
Req = 4 Ω
Hence, the total circuit resistance is 4 Ω.
Effects of Heat on the Conductor
The heating effect may cause an increase in the temperature of the wire of the conductor.
This may also cause an increase in the volume of the material.
More simply, when an electrical current is passed through a conductor, it generates excess heat due to the resistance caused by the electrons in the conductor to the flowing current. The work done in overcoming this resistance to the current generates what we call heat in that conductor. The electrical heating effect of the electrical current is most commonly and widely applied and used in our daily life. For example, electrical irons, kettles, toasters, electrical heaters, etc. are used widely as alternatives to conventional methods of cooking and also laundry. This same effect is used widely in electrical bulbs which are alternatives to conventional incandescent lamps. These devices have modernized and revolutionized the new sustainable world over the years.
Let us assume a current I that is flowing through a resistor that has a resistance of R as shown in the circuit. Let the potential difference across ends of the terminals of the battery be V. Let us assume to be the time during which a charge of Q amount flows across the circuit. The work which is done in moving that charge Q through a potential difference V is V × I.
Hence, the source has to supply energy equal to V × I in time t. Therefore, the power input to the electrical circuit by the source is
P = V × Q/t
= V × I
Or the energy that is being supplied to the circuit by the source in time t is P × t, that is, V × I × t. This extra energy generated gets dissipated in the resistor in the form of heat. Therefore, for a steady and fixed current I, the amount of heat denoted by H that is produced in time t is
H = V × I × t
Joules Law of Heating
The very famous physicist James Prescott found that the amount of heat generated per second that develops in a conductor having a current is directly proportional to the electrical resistance of the wire and also with the square of the current given. This heat which is liberated or generated because of the electrical current that flows in an electrical wire is expressed in Joules.
By applying the ohms law to the equation H = V × I × t. We can deduce the Joules law or joules first law which gives the relationship between the heat that is produced by flowing charges of electric current through a conductor. It is directly proportional to the square of the supplied current, the electrical resistance exerted by the appliance, and the time for which we used it. This is known or called joule’s law of heating. The following is its expression:
H = I2 × R × t
where,
H gives or indicates the amount of heat.
I shows the amount ofthe electrical current supplied.
R is the value or amount of electric resistance exerted in the conductor.
t denotes the time for which the appliance is operated.
Factors on which Heat Depends
The amount of liberated or generated heat is directly proportional to the given wire’s electrical resistance when the electrical current in the given circuit and the flow of supplied current is not altered or changed.
The amount of liberated or generated heat in the conductor carrying current is directly proportional to the square of the electrical current that flows through the given circuit when the electrical resistance and current supply is kept constant.
The amount of heat generated or produced because of the electrical current flow is directly proportional to the time of usage of flow when the electrical resistance and the current flow is constant.
Applications of Heating Effect of Electric Current
Following are some of the widely used common devices in which the heating effect of current is used and harnessed for other purposes:
Electric Fuse
In any electrical instrument which we sometimes use due to a sudden rise in the amount of current, the instrument or appliance gets overheated or burnt down which sometimes may result in a severe fire. A conducting wire with a very low melting point is joined or connected in a series connection with the device or appliance circuit to avoid any mishap or this type of accident. Whenever the current value somehow accidentally rises, the wire inside the fuse melts due to the excessive heating and thus results in breaking the electrical circuit saving the device as well as our lives. We choose or select the fuse according to the appliance used. A device or an appliance that works on a higher current needs a greater value of fuse and vice versa.
Electric Bulb
Electrical bulb contains a very thick metallic wire which is in turn made up of highly resistive tungsten metal. This metal is always kept in an inert environment so that it doesn’t react with a neutral gas or vacuum. When the electrical current flows through the used tungsten wire, it becomes warm or heated, and then it emits light. Most of the electrical power which is drawn in the electrical circuit from the electrical source is liberated or dissipated in the form of heat and the rest is given or emitted in the form of light energy. The tungsten filament used also has a high resistivity and a very high melting point so that it doesn’t get heated easily when used.
Electric Heater
In an electrical heater, a very high resistance nichrome wire is mostly and commonly used as a coil. The coil is rotated or wound on grooves which are made up of the ceramic material of the iron plate or china clay plate. Whenever the electrical current flows in the coil, it quickly becomes warm or heated, which is then widely used to heat our cooking vessels. In mountain areas, electrical room heaters are used to keep their rooms warm and heated to save themselves from the exhaustive cold outside.
Electric Iron
Between the metal part and the electrical coil in an iron, Mica is placed which is by nature an insulator. The coil of the iron becomes warm or heated with the continuous passage of current which is then passed on or transferred to the metallic part through the mica used. Finally, after a while, the metallic part becomes very heated or whatever temperature we have set, which is then used for ironing different material clothes according to our wish.
For using and harnessing the heating effect of electric current, the element of appliances needs or is required to have a high melting point to retain more heat.
Solved Examples on Heating Effect of Electric Current
Example 1: 100 J of heat energy is produced by an electrical appliance that is used for 1 sec and is having a resistance of 4 ohms. Find and calculate the potential difference of the appliance.
Solution:
As we know,
H = I2×R×t
Given, H = 100 J, R = 4 ohm, t = 1 sec
Therefore, 100J=I2×4×1
I = 5 A
From ohm’s law we know
V=I×R
Therefore, V = 5A × 4Ω
V = 20 V
Therefore, the potential difference generated is 20 V.
Example 2: Find out the heat produced by the electric toaster when it is used for 5 minutes. The current given was 2 A and its resistance is 3 ohms.
Solution:
As we know,
H = I2×R×t
Given, I = 2 A, R = 3 ohm, t = 5 min = 300 sec
Therefore, H=22×3×300J
H = 22×900 J
Therefore, H=4×900 J
H = 3600 J
Hence, the heat that is produced or liberated by the electrical toaster is 3600 J.
Example 3: The heat generated is 100 J from an electrical fan which has a potential difference of 10 V and the time for which it is used is 10 sec. Find what the amount of electrical current that is used is?
Solution:
As we know,
H = V × I × t
Given, H = 100 J, V = 10 V, t = 10 sec
Therefore, 100J=10V×I×10
∴100J = 100 × I
I = 1 A.
Therefore, the current used is 1 A.
Example 4: What is the power consumed if a device or appliance is operated at 1 V potential difference and 6 A current?
Solution:
As we know,
P = V × I
Given, V = 1V AND I = 6 A
P = 1 V × 6 A
P = 6 W
Hence, the power that has been consumed is 6 W.
Example 5: What is the power consumed if a toaster having 3-ohm resistance is operated at 1 A current?
Solution:
As we know,
P = I2 × R
Given, I = 1 A and R = 3 ohm
∴P = 12 × 3Ω
P = 3 W.
Therefore, the power consumed has been calculated to be 3 W.
FAQs on Heating Effect of Electric Current
Question 1: What is the Heating Effect of Electric Current?
Answer:
A high resistance wire, such as nichrome wire, heats up and emits heat when an electrical current is carried through it. The heating action of current is what causes the Heating effect of Current.
Question 2: Name two Appliances based on the Heating Effect of Electric Current.
Answer:
Two appliance that work on the principle of Heating Effect of Electric Current are Electric Fuse and Electric Iron.
Question 3: Mention two Practical Disadvantages of the Heating Effect of Electric Current.
Answer:
Two Practical Disadvantages of the Heating Effect of Electric Current are:
A part of the electrical current that travels through the conductor is transformed to heat. Often, this is a waste of energy.
The heat generated can start a fire or harm the insulation and other electrical parts.
Question 4: Write a common example of the Heating Effect of an Electric Current.
Answer:
Electrolysis is an important and common example of the heating effect of electric current.
Question 5: What is the Heating Effect of the Electric Current Formula?
Answer:
The Heating Effect of Electric Current Formula is given by,
H = I2 × R × t
where,
H gives or indicates the amount of heat.
I shows the amount ofthe electrical current supplied.
R is the value or amount of electric resistance exerted in the conductor.
t denotes the time for which the appliance is operated.
Commercial Units of Energy
Here, we must define the commercial unit of electric energy and explain:
How it differs from the home unit of electric energy.
A kilowatt-hour is equal to one joule.
In a commercial setting, such as factories or other business sectors, how electrical equipment utilizes one kilowatt of energy for one hour.
Kilowatt-hour (kWh) is a commercial unit of energy. A kilowatt-hour is an amount of energy consumed by a machine working at a steady rate of one kilowatt for one hour. As a result, Joule is a SI unit of energy, but Kilowatt-hour is a Commercial unit.
Electrical Energy
The energy generated from the electric potential or kinetic energy of charged particles is known as electrical energy. It’s the energy that’s been transformed from electric potential energy, in general. The energy created by the passage of electrons from one location to another is known as electrical energy. Current or electricity is the flow of charged particles along/through a media (such as a wire).
Relation between the commercial unit and SI unit of energy
Kilowatt-hour gives a direct relationship between energy and power. Electric power in terms of energy is defined as the amount of electrical energy consumed per unit of time, i.e., electrical energy divided by time is equal to electrical power. SI unit of power is Watt and represented by W.
The formula of power in terms of energy is given as:
P = E ⁄ t
Here,
P is the power generated
E is the energy consumed
t is the time
Rearrange the formula in terms of energy.
E = P t
Representing the Kilowatt-hour in terms of above formula.
Consider the power equal to 1 kW and time equal to 1 h.
E = P t
1 kW h = 1kW × 1 h
1 kW h = 1000 W × 1 h
1 kW h = 1000 (J / s) × 3600 s
1 kW h = 3.6 × 106 J.
Therefore, the kilowatt-hour (kW h) is a unit of energy equal to 1 kilowatt of power expanded for 1 hour. It does not refer to the number of kilowatts you consume every hour. It’s just a unit of measurement that represents the amount of energy used if a 1000 Watt appliance was left on for one hour. It would take 10 hours to accumulate 1 kW h of energy if you switched a 100 Watt light bulb.
Although the kilowatt-hour is not a recognized system unit, it is widely used in electrical applications. The unit of electrical energy is the kilowatt-hour (kWh), which is the same for commercial and household use. Commercial apartments have distinct rates of charge, which are higher than residential unit charges. It is mostly used to reduce power usage in factories and other industries that use a lot of electricity.
Note: The kilowatt-hour is the commercial unit of electric energy here, and 1 kW h = 3.6 × 106 J. So, be cautious while converting kilowatt-hours to joules.
Sample Problems
Problem 1: Calculate the number of Joules contained in 1 unit of energy.
Solution:
1 unit of energy is equal to 1 kW h.
1 kW h = 1000 W × 1 h
=1000 (J / s) × (60 × 60) s
= 3600000 J
= 3.6 × 106 J.
Hence, the number of Joules in 1 unit of energy is equal to 3.6 × 106 J.
Problem 2: Why does the Kilowatt-hour is used as the commercial unit of energy?
Solution:
In practice, the unit kWh is used for the measurement of electrical energy, rather than joule. This is because Joule is a very small unit and the energy consumption in day to day life is very large, i.e., it comes in figures of 106 to 108. Thus, to reduce the complexity of handling such large figures, a bigger unit was required. This bigger unit used for the measurement of electrical energy is kWh and is related to Joule as: 1kWh = 3.6 ×106 J. Hence, the energy reading commercially became simpler by using this bigger unit instead of joule
Problem 3: Name the unit used in selling electric energy to consumers. An electric motor takes 10 A from a 110 V line. Determine the power of the motor and energy consumed in 3 hrs. Calculate the cost of electric energy for the month of June at a rate of Rs. 9 / unit.
Solution:
Kilowatt-hour (kWh) unit is used in selling electric energy to consumers.
Given:
Current drawn to motor, I = 10 A
Voltage across the motor, V = 110 V
Time for energy is consumed, t = 3 h
Power generated by motor, P = V I
= 110 V × 10 A
= 1100 W
Energy consumed by motor, E = P × t
= 1100 W × 3 h
= 3.3 kW h
Now,
Energy consumed, E = 3.3 kW h
Cost of energy per unit, c = Rs. 9 ⁄ unit
Number of days, n = 30
Total Cost, C = E c n
= Rs. (3.3 × 9 × 30)
= Rs. 891.00
Hence, selling unit of energy is Kilowatt-hour, power generated by motor is 1100 W, energy consumed by motor is 3.3 kW h, and cost of energy is Rs. 891.
Problem 4: Convert 72000 J of energy into kilowatt-hours.
Solution:
Since, 3.6 × 106 J = 1 kW h
So, 1 J = (1 ⁄ 3.6) × 10-6 kW h
Therefore, 72000 J = (72000 ⁄ 3.6) × 10-6 kW h
= 2 × 10-3 kW h
Hence, 72000 J of energy is equal to 2 × 10-3 kW h.
Problem 5: Differentiate between watt and watt-hour.
Solution:
Watt is the unit of power while watt hour is the unit of work/energy, since energy = power × time.
Problem 6: An electrical appliance consumes 7.2 MJ of energy to generate power equal to 2 kW. Find the time in minutes for which the given energy is consumed.
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