NCERT Notes Class 10 Science (Physics) Chapter 11 Electricity - Param Himalaya

NCERT Notes for Class 10 Science Physics Chapter 11 Electricity is prepared and uploaded for reference by a team of expert members of Param Himalaya. Get NCERT Exercise Solution Class 10 Science Chapter 11 Electricity.

Electricity : 

Electricity is a controllable and convenient form of energy and defined as the flow of electric charge. 

Electric Charge  : 

Electric charge can be defined as a fundamental property of subatomic particles that gives rise to the phenomenon of experiencing force in the presence of electric and magnetic fields.

Some  properties of electric charge

1. Electric charges are of two types - Positive charges or protons have a charge of $+1.6 × 10^{-19}$ Coulomb. Negative charges or Electrons have a charge of $-1.6 × 10^{-19}$ Coulomb.

2. Charge is a scalar quantity.

3. Charge is transferable, Charge( Electron) transfer from one body to another.

4. Like charges repel each other and unlike. charges attract each other.

Unlike charge attract each other - param Himalaya

Unlike charge attract each other - param Himalaya

4. Charge is always associated with mass.  
5. Net charge on Object q = ne , Where, n = Integer numbers , e = value of the charge $(1.6× 10^{-19}$ C)

Current

The rate of flow of electricity i.e ., The quantity of charge flowing per unit time , is called Current . 

Electric Current - Param himalaya


If in t seconds q change passes through a point of conductor.

Then charge flowing in 1 second = $\frac{q}{t}$

Or I = $\frac{q}{t}$

Or q = I×t

If q = 1 Coulomb , t = 1 Second 

Then I =  $\frac{ 1 Coulomb}{1 second}$ 

I = 1 A

The unit of current intensity is Ampere , it is represented by "A".

Ammeter : 

  1. It is an electrical device that is used to measure the intensity of current in a circuit
  2. Ammeter is always connected in series.
    Ammeter - Param Himalaya
    3. Ammeter has Low resistance.

Voltage or Electric Potential Difference :

The electric potential difference between two points in an electric circuit carrying some current as the work done to move a unit charge from one point to the other.

Voltage or Electric Potential Difference - Param himalaya

Potential difference (V) between two points = $\frac{Work done (W)}{Charge (q)}$

V = $\frac{W}{q}$

The unit of Electric Potential difference is Volt , it is represented by "V".

If q = 1 Coulomb , W= 1 Joule 

Then V =  $\frac{ 1 Joule}{1 Coulomb }$

V = 1 Volt

Voltmeter :

  1. It is an electrical device that is used to measure the potential difference between any two points.
  2. Voltmeter is always connected in parallel.
    Voltmeter - Param Himalaya
    3. Voltmeter has High resistance.

Electric Circuit : 

An electric circuit is a continuous and closed path in which electrons move to produce electric currents. The components included in an electric circuit are a battery, connecting wire, switch and any electric load.

Electric Circuit - Param Himalaya


Symbols of Some commonly used components in circuit diagrams : 


Ohm's Law : 

The current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor while temperature remains constant and understanding the influence of no magnetic field.

Thus , if I is the current flowing through a conductor and V is the potential difference ( or voltage ) across the conductor.

$V \propto I$

V = IR

Ohm's Law - Param Himalaya

Where R is a constant called the resistance of the conductor.

Ohm's Law Magic Triangle: 

The Ohm's law Triangle is actually a visual representation of Ohm's law which makes it easy to understand it. Let's consider a triangle with three labels V,I and R. 

Ohm's Law Magic Triangle - Param Himalaya

The Above triangle can easily be used to obtain three equations. 

$$V = IR$$

$$I = \frac{V}{R}$$

$$R = \frac{V}{I}$$

Draw a V-I Characteristics or graph for a conductor obeying Ohm's law.

V-I Characteristics graph - Param Himalaya

V 0.4 0.8 1.2 1.6
I 0.1 0.2 0.3 0.4

slope of V-I graph represents the resistance of the conductor.

$$Slope = \frac{V}{I}$$

$$R = \frac{1.6-0}{0.4-0} = 4\Omega$$

Limitation of Ohm's Law : 

  1. All conductors do not obey Ohm's law.
  2. Temperature of the conductor should not change.
  3. Ohm's law is not applicable to vacuum tubes such as diodes , triodes etc.
  4. It is not applicable to gaseous conductor such as gases in discharge tube.

Resistance of Conductor : 

The property of conductor by virtue of which it opposes the flow of electric current through it is known as resistance of the conductor.

(i) Resistance is denoted by the letter R.

(ii) The SI unit of resistance is ohm. the ohm is denoted by the Greek letter ( $\Omega$) called omega.

(iii) Resistance is a Scalar quantity.

Resistance of a Conductor depends on the following factors :

(i) Length of the conductor (l).

(ii) Area of cross-section of the conductor ( or Thickness of the conductor) (R).

(iii) Nature of the material of the conductor.

Resistance of a Conductor depends on the following factors

On the basis of Experiments that :

(i) The resistance of a given conductor is directly proportional to its length.

$R \propto l$

(ii) The resistance of a given conductor is inversely proportional to its area of cross-section.

$R \propto \frac{1}{A}$

then , 

$R \propto \frac{l}{A}$

$R = \frac{\rho \times l}{A}$

Where $\rho$ (rho) is a constant known as resistivity of the material of the conductor. Resistivity is also known as Specific resistance.

Resistivity : 

The resistance (R) of a conductor is directly proportional to its length (l) and inversely proportional to its area of cross - section (A) , so we have 

$R \propto \frac{l}{A}$ 

or $R \propto \rho\frac{l}{A}$

Where $\rho$ (rho) is called resistivity of the material of conductor.

Resistivity of a conductor - Param Himalaya

if , l =1 m and A = 1 $m^{2}$

then R = $\rho$

$\therefore$ Resistivity of a conductor is equal to resistance of the conductor if its length is 1m and area of cross section 1$m^{2}$.

Unit of Resistivity :

we know : 

$R \propto \rho\frac{l}{A}$

$\rho = \frac{R \times a}{l}$

So, SI unit of resistivity ($\rho$) = $\frac{ohm \times m^{2}}{m}$ = ohm-m

Thus , SI unit of resistivity is ohm-m (or $\Omega$.m)

Combination of Resistors : 

Resistors are used in various combinations. There are two methods of arranging the resistors in different combinations:

(i) Resistors in Series

(ii) Resistors in Parallel

Equivalent Resistance: The equivalent resistance of combinations of resistors in series is equal to the sum of their individual resistances in the circuit.

Resistors in Series Combination: 

Two or more resistances are said to be connected in series when they are connected end to end.

In this case, the equivalent or the total resistance equals the sum of the number of individual resistances present in the series combination.

Consider a case of three resistances ($R_{1}$, $R_{2}$, and $R_{3}$) connected in series with each other with the corresponding voltage source ($V_{1}$, $V_{2}$ , and $V_{3}$ ) in a circuit shown below:

Resistors in Series Combination - Param Himalaya


The equivalent current flow through it is I, measured 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.

$V=V_{1}=V_{2}=V_{3}$

The current I through each resistor is the same ( For single path) i.e. 

$I=I_{1}=I_{2}=I_{3}$

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:

$V_{Eq}$= $IR_{Eq}$

By applying Ohm’s law to all resistors individually as:

$V_{1}$= $IR_{1}$

$V_{2}$= $IR_{2}$

$V_{3}$= $IR_{3}$

Hence,  

$IR=IR_{1}+IR_{2}+IR_{3}$

$R_{Eq}=R_{1}+R_{2}+R_{3}$


Conclusions

1. The current through the circuit will remain the same here.

2. The equivalent potential difference is the sum of the individual potential difference across each resistor.

3. As a result, equivalent resistance becomes the sum of individual resistances.

4. 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. 

5. The series combination is needed to increase the resistance and to divide high potential differences across many resistances.

6. 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. 

Consider a case of three resistances ($R_{1}$, $R_{2}$, and $R_{3}$) connected in parallel with each other with the corresponding voltage source ($V_{1}$, $V_{2}$, and $V_{3}$) in a circuit shown below:

Resistors in Parallel Combination - Param Himalaya


Here, the current flows through each resistor is different therefore, the equivalent current flown through the circuit is:

$I_{Eq}=I_{1}+I_{2}+I_{3}$

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:

$I_{Eq}=\frac{V}{R_{Eq}}$

On applying Ohm’s law to individual resistors as:

$I_{1}= \frac{V}{R_{1}}$

$I_{2}= \frac{V}{R_{2}}$

$I_{3}= \frac{V}{R_{3}}$

$\frac{V}{R}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}$

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

In conclusion.

1. 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.

2. The equivalent current through the circuit is the sum of individual currents through each branch of the circuit.

3. The potential difference across the two terminal points of the circuit remains the same.

4. the reciprocal of equivalent resistance of the circuit is the sum of reciprocal of the individual resistances.

5.In a parallel circuit, a resistor or some other component can be easily connected or disconnected without disturbing the other components.

6. In parallel combination, the current flown in the circuit is divided into different branches and hence each component receives the required amount of current.

7. the equivalent resistance is always lesser than all the individual resistances.

8.If one of the components fuse or shorted, the rest of the components of the circuit works usually.

Work and Power : 

When current is passed through a resistor , it offers a resistance and work must be done continuously to maintain potential difference between two point of the resistor and if 'W' work is done , if charge 'Q' is brought form one point from one point to other and potential is V.

then 

$V = \frac{W}{Q}$

or $W = V \times Q$

but $Q = I \times t$

$W = V \times I \times t$

Power : 

Rate of doing work or work done in 1 Second , is called the power . It is denoted by "P".

$Power (P) = \frac{Work done (W)}{Time (t)}$ 

$P = \frac{V \times I \times t}{t}$

$P = V \times t$

Unit of Power is Joule / second or J/s which is also called "Watt" (W).

Since work done is equal to amount of energy consumed.

hence , 1 Watt = 1 volt \times 1 ampere

Watt is very Small unit of power , in practice a large unit "Kilowatt" or "KW" is used and

1 kW = 1000W

Other Formulae of energy consumed  :

$W or E = V \times I \times t$
$W = IR \times I \times t (\because V = IR)$
$W = I^{2} \times R \times t$
and $E or W = V \times I \times t$
$E= V \times \frac{V}{R} \times t (\because I = \frac{V}{R})$
$E= \frac{V^{2}t}{R}$
So, $W or E = VIt pr I^{2}Rt or \frac{V^{2}t}{R}$
similarly , $P = VI or I^{2}R or \frac{V^{2}}{R}$

Energy Consumed 
We know that , $Power = \frac{Work or Energy}{time}$
or \frac {Power} {times} = Energy
If power is in watts(W)and time is in hours (h)
Then $energy consumed = Power \times time $
$= Watt \times hour $
$= Watt \times h$
= Wh 
So , 'Watt hour' is the unit of electrical energy , and 
1 Wh = 1 W \times 1h
Means if an appliance of power 1Watt is used for 1 hour , Energy consumed is 1 watt-hour (Wh)
If Power = 1kW, Time = 1 hour
Then , energy consumed =1 kW \times 1h = 1kWh
or $1000 W \times 1 h = 1000 Wh$
So, 1kWh = 1000 Wh
or 'Kilowatt hour is the commercial unit of energy . It is one unit of electrical energy'.

Conversion of kWh into joule 
$1 kWh = 1kW \times 1 hour$
$= 1000 Watt \times 60 \times 60 second$
$= 3600000 joule = 3.6 \times 10^{6} joule.$

Heating Effect  of Current :

When an electric current is passed through a conductor , it generates heat due to the hindrance caused by the conductor to the flowing current. The work done in overcoming the hindrance to the current generates heat in that conductor.

Heating Effect of Electric Current Formula :

Let an electric current I is flowing through a resistor having resistance equal to R. The potential difference through the resistor is equal to V.

The charge Q flows through the circuit for the time t. Thus , work done in moving of charge Q of potential difference V.

$V = \frac{W}{Q}$

when W = H

$V = \frac{H}{Q}$

H = VQ ------- (i)

we know that 

Electric current $I = \frac{Q}{t}$

Q = It

then H = VIt ---------- (ii)

from ohm's law 

V=IR so,

$H = IR\times It$

$H = I^{2}Rt$--------- (iii)

This is joules law of heating. SI unit of Heat is Joule (J).

Factors on Which Heat Depends :  

(i) Directly proportional to the square of the current.

$$H \propto I^{2}$$

(ii) Directly proportional to the resistance of a given current.

$$H \propto R$$

(iii) Directly proportional to the time for which the current flows through the resistor.

$$H \propto t$$

so , $$H \propto I^{2}Rt$$ 

$H = I^{2}Rt$ Joules law of Heating.

Application of Heating Effect of Electric Current :

(i) Alloy are used in making elements of heating devices : 

Resistivity of an alloy is much higher than the metals which is /are present in its composition. For example, resistivity of copper is $1.62 \times 10^{-8}$ ohm m. while the resistivity of its alloy manganin (Cu 84% , Mn 12% , Ni 4%) is $4.44 \times 10^{-5}$ ohm m and the resistivity changes at very low rate with change in temperature and moreover alloy do not oxidise (burn) on high temperature. this is the main reason alloy such as manganin and constantan are used to make resistance wire in electronic appliance. such as Electrical iron , electrical press , electrical oven , room heater , electric kettle , water heater etc. mostly, nichrome and alloy is used to make the heating element of these devices. The resistance of nichrome is very high and it becomes red hot on passing current due to generation of heat.

(ii) Thin wire of tungsten is used for making the filament of incandescent bulbs : 

Because its melting point is very high ($3380^{0}C$) .We know that resistance of a resistor is inversely proportional to its area of crosssection. So to increase its resistance, its area of cross section is decreased or a thin wire is used for making filament of bulb. On passing current it becomes red-hot and emits light. , its temperature is about $2500^{0}C$. That's why , an alloy or metal such as tungsten is used which has melting point about $3380^{0}C$. Moreover, at high temperature ($2500^{0}C$), it may react with atmospheric oxygen (air) or can oxidise (burn). So , in incandescent bulb air is evacuated or it is filled with non-reactive gas such as nitrogen or argon. Most of the current consumed in an electric bulb is used to produce heat and only a small amount of it changes in the form of light.

(iii) Use of heating effect of current in electric fuses : 

A fuse wire is a thin wire of a metal or an alloy of tin and lead , having low melting point . As it is thin (having less area of cross-section) its resistance is higher than wires used in domestic electric circuit. When current of higer voltage (due to any reason) passed through the fuse wire, it becomes hot and melts and breaks the circuit and flowing of current stops and thus protects the electrical appliance used in the domestic circuit from damage. In domestic circuits fuse wire of 1A to 5 A are used.


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