Check here the revision notes for CBSE Class 10 Science Chapter 12 - Electricity. Subject experts at Jagran Josh have prepared these chapter notes to facilitate quick revision before the upcoming CBSE Class 10 Science Board Exam 2021. The best thing is about these notes is that these have been prepared based on the revised CBSE Class 10 Science Syllabus only. So, students can revise all the topics prescribed by the board quickly within a few minutes and in the spare time they can practice with previous years' question papers to score well in the exam.
Revision Notes for CBSE Class 10 Science Notes for Chapter 12 - Electricity:
Electric current: The rate of flow of electric charge is known as Electric Current. Electric current is carried by moving electrons through a conductor.
If a net charge Q, flows across any cross-section of a conductor in time t, then the current I, through the cross-section is represented as:
The SI unit of electric current is Ampere (A).
Ampere: One ampere is equal to the flow of one coulomb of charge per second, that is,
1 A = 1 C/1 s
Potential Difference:Electric potential difference is defined as the work done to move a unit charge from one point to the other.
Potential difference (V) between two points = Work done (W)/Charge (Q) V = W/Q
The SI unit of electric potential difference is volt (V).
1 Volt: One volt is the
When 1 joule work is done in carrying one Coulomb charge from one point to another then potential difference between two points is called 1 volt.
Thus, 1 V = 1 J C–1
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Ohm’s law: It states that the potential difference across the two points of a metallic conductor is directly proportional to current passing through the circuit provided that temperature remains constant.
V ∝ I
or V/I = constant = R
or V = IR
where, R is called resistance and is a constant.
Resistance: The property of a conductor to resist the flow of charges through it is called resistance of he conductor.
Resistance of a conductor depends on:
(i) length of conductor.
(ii) area of cross-section of conductor.
(iii) nature of material of conductor.
Resistivity (ρ): The electrical resistance offered by a substance of unit length and unit cross-sectional area is called resistivity.
The SI unit of resistivity = Ω m
Resistivity does not change with change in length or area of cross-section but it changes with change in temperature.
Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc.
Copper and aluminium are used in electrical transmission due to their low resistivity.
Series combination of resistors: When two or more resistors are connected end to end, they are said to form a series combination.
Parallel combination of resistors: When two or more resistors are connected between two common points, they are said to form a parallel combination.
Series Combination of Resistors | Parallel Combination of Resistors |
Current remains the same through all resistors. | Voltage remains same through all resistors. |
Total Voltage, Vs = V1 + V2 + V3 | Total current, Ip = I1 + I2 + I3 |
By applying Ohm’s law, V = I R V1 = I R1 V2 = I R2 and V3 = I R3 ⇒ I R = I R1+ I R2 + I R3 Or Rs = R1 +R2 + R3 Thus, equivalent resistance of the combination is equal to the sum of individual resistances. | By applying Ohm’s law, I = V/Rp I1 = V /R1 I2 = V /R2 and I3 = V /R3 ⇒ V/Rp = V/R1 + V/R2 + V/R3 Or 1/Rp = 1/R1 + 1/R2 + 1/R3 Thus, reciprocal of equivalent resistance is equal to sum of reciprocals of individual resistances. |
Advantages of parallel combination over series combination:
(i) In series circuit, when one component fails the circuit is broken and none of the component works.
(ii) Different appliances need currents of different values to operate properly. Therefore, they cannot be connected in a series combination as current remains same in the series circuit.
(iii) The total resistance in a parallel circuit is decreased.
Heating effect of electric current: If an electric circuit is purely resistive then the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current.
For example: When electric energy is supplied to an electric bulb, the filament gets heated because of which, it gives light.
Cause of heating effect of electric current: Electric current generates heat to overcome the resistance offered by the conductor through which it passes. Higher the resistance, the electric current will generate higher amount of heat.
Joule’s law of heating: It states that the heat produced in a conductor is directly proportional to
(i) square of current for a given resistance,
(ii) resistance for a given current, and
(iii) time for which the current flows through the resistor
Thus, H = I2Rt
Application of heating effect of electric current:
(i) In an electric bulb, the filament of bulb gives light because of the heating effect of electricity. The filament of bulb is generally made of tungsten metal because it has a very high melting point and also it does not oxidize readily at a high temperature.
(ii) Electric fuse is a safety device to protect the electrical appliance from short circuit. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit saving the appliances attached in the circuit.
(iii) The element of electric iron is made of alloys having high melting point. Electric toaster, electric oven, electric heater and geyser work on the same mechanism.
Electric power (P): The rate at which electric energy is dissipated or consumed in an electric circuit, is termed as electric power.
P = VI
Or P = I2R = V2/R
The SI unit of electric power is watt (W).
One watt is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.
The commercial unit of electric energy is kilowatt hour (kW h).
1 kW h = 1000 watt × 3600 second = 3.6 × 106 watt second = 3.6 × 106 joule (J)
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