Electricity

Electricity can be divided into two parts : static electricity and current electricity. Most practical applications of electricity are based on current electricity. "An electric current consists of charges in motion from one region to another."

1.0Introduction

We study electricity into two parts: (1) Static electricity, deals with charges at rest and phenomenon associated with them. (2) Current electricity, deals with charges in motion and phenomenon associated with them.

2.0Static electricity (or frictional electricity)

Most of the substances release static electricity when rubbed against another. This is most noticeable when the rubbed substance is a very good insulating material.

To The electricity developed on the surfaces of insulating bodies when rubbed against each other is called frictional electricity or static electricity. Source of static electricity When two substances are rubbed together, some electrons are removed from the atoms on the surface of one and transferred to the other. The substance which gains electrons becomes negatively charged and the one which loses electrons becomes positively charged. Thus, the excess or deficiency of electrons makes a substance negative or positive respectively (see figure).

3.0Electric charge

Electric charge is the property associated with matter due to which it produces and experiences electric and magnetic effects.

List of objects acquiring two kinds of charges on rubbing

4.0Properties of electric charge

(1) Electric charge is a scalar quantity. It can be of two types : positive and negative. (2) Like charges repel each other while unlike charges attract each other. [see figure (a) and (b)]

Examples: (a) Rubbing a glass rod on silk cloth results in both becoming charged, one positive and another negative. The total charge remains zero, as it was in the beginning. This means that frictional electricity is the transfer of electric charge from one body to the other, and not the creation of charge. (b) Conservation of electric charge holds good in all types of reactions either 'chemical' or 'nuclear' reactions. (5) Quantisation of charge : Every charge is an integral multiple of a certain smallest amount of charge that exists in nature. This property of charge is called 'quantisation'. $\text{ Q = }\pm \text{ ne }$ Where, $\mathrm{n}=1,2,\mathrm{3.........}$. and e is called 'elementary charge', the smallest amount of charge that exists independently in nature. $e=1.6\times 10^{-19}\mathrm{C}$, ' e ' is the magnitude of charge of an electron (or proton). Reason : Since, electrons are indivisible, thus, only integral number of electrons can be transferred from one body to another, on rubbing. Hence, the charge bodies will have charges which are integral multiples of the charge on electron (e). (T) Electric charge of an electron, $\mathrm{q}=-\mathrm{e}=-1.6\times 10^{-19}\mathrm{C}$ (8) Electric charge of a proton, $\mathrm{q}=+\mathrm{e}=+1.6\times 10^{-19}\mathrm{C}$

Building Concepts 1 If a neutral body is made positively charged, is there any change in its mass? Explanation: In charging any neutral body, the mass of a body changes, though the change is extremely small or negligible. If a neutral body is made positive, it means electrons are removed from it. Thus, the mass of body decreases.

Unit of electric charge SI unit : Coulomb (C) CGS unit : Electrostatic unit (esu); also called statcoulomb (stat C) 1 Coulomb $=3\times 10^9$ esu Other units of electric charge : (1) Faraday (F) : $1\mathrm{F}=96500\mathrm{C}$ (2) Ampere-hour(A-hr) : $1\mathrm{A}-\mathrm{hr}=3600\mathrm{C}$

Definition of 1 coulomb (in terms of electric current) One coulomb is defined as the amount of charge that flows through a given cross-section of a wire in one second if there is a steady current of one ampere flowing in the wire. $1\mathrm{C}=1$ A.s $(\because \mathrm{Q}=\mathrm{I}\times \mathrm{t})$

Number of electrons in 1 coulomb According to quantisation of charge, $\mathrm{Q}=$ ne or $\mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}=\frac{1\mathrm{C}}{1.6\times 10^{-19}}=6.25\times 10^{18}\approx 6\times 10^{18}$ electrons

5.0Electric potential

Electric potential at a point A in an electric field is the work done per unit positive charge in transporting it from infinitely far away to the point A. (see figure) $\mathrm{V}=\frac{\mathrm{W}}{{\mathrm{Q}}_0}$

Electric potential is a scalar quantity. It can be positive or negative.

Unit of electric potential SI unit : Volt (V) $1\text{ Volt = }1\text{ Joule/coulomb }=1\mathrm{J}{\mathrm{C}}^{-1}$

Definition of 1 volt in terms of electric potential 1 volt is the electric potential at any point A when 1 joule of work is done in moving a charge of 1 coulomb from infinity to the point $A$.

Potential difference Potential difference between two points is defined as the work done in carrying a unit positive charge from one point to another point. (see figure)

Potential difference between two points $\mathrm{V}=\frac{\mathrm{W}}{{\mathrm{Q}}_0}$

Definition of 1 volt in terms of potential difference 1 volt is the potential difference between two points when 1 joule work is done in moving a charge of 1 coulomb from one point to another.

• If work done in moving a unit positive charge from point $A$ to point $B$ is zero, it means, potentials at point $A$ and point $B$ are same, i.e., $V_A=V_B$.
• Potential difference between two points is independent of actual path followed between the points.

6.0Conductors and insulators

Conductors are the substances in which electricity can flow quite freely. $◯$ The substances which allow the electric current to flow through them are called 'conductors'.

Examples of conductors : (1) Metals : Copper (Cu), aluminium (Al), silver (Ag), iron (Fe), etc. (2) Non metals : Graphite (3) Electrolytic solution : Aqueous solutions of $\mathrm{NaCl},\mathrm{HCl},{\mathrm{H}}_2{\mathrm{SO}}_4$, etc. $◯$ The substances which do not allow the electric current to flow through them are called 'insulators'.

Examples of insulators : Rubber, glass, wood, plastic, porcelain, pure water, sugar, etc.

7.0Current electricity

The study of charges in motion and their effects is called 'current electricity'. The rate of flow of electric charge through any cross-section is called 'electric current'.

$\mathrm{I}=\frac{\mathbf{Q}}{\mathrm{t}}$ Where, ' $Q$ ' is the charge that flows across the cross-section area of a conductor in time ' $t$ '. Electric current is a scalar quantity. Though, a direction is associated with electric current, still it is not considered as vector quantity because it does not obey the vector laws but obeys scalar laws of addition.

Unit of electric current

Electric current obeys scalar laws of addition The electric current is expressed by a unit called ampere (A), named after the French scientist, Andre-Marie Ampere (1775-1836). Small quantities of current are expressed in milliampere or in microampere.

SI unit : Ampere 1 Ampere $=1$ coulomb $/\sec =1{\mathrm{Cs}}^{-1}$ $\star 1\mathrm{mA}=10^{-3}\mathrm{A}\star 1\mu \mathrm{A}=10^{-6}\mathrm{A}\star 1\mathrm{ms}=10^{-3}\mathrm{s}$

Definition of 1 ampere 1 ampere is the electric current flowing through a conducting wire when 1 coulomb charge flows through it in 1 second.

Conventional direction of current The conventional direction of electric current is 'the direction of flow of positive charge'. This means direction of electric current is 'opposite to the flow of negative charge'.

Current carriers The charged particles which flow in a particular direction to produce electric current are called 'current carriers'.

Comparison between water current and electric current

In solid conductors, current carriers are 'free electrons'. In liquids, current carriers are positive ions and negative ions. In gases, positive ions and electrons are current carriers.

Electromotive Force (emf)

Electric current is possible in a closed circuit if there is a source of external force which compels the current to move in a definite direction. Some sources of emf are: (1) Battery or electrochemical cell (2) Electric generators (3) Solar cells.

The source of emf maintains a potential difference (V) between its terminals by doing work, which compels the electric current to flow in a particular direction in a closed circuit. (see figure) $\mathrm{Emf}(\epsilon )$ : It is the maximum work done by a source of emf in taking a unit positive charge once around a closed circuit. $\epsilon =\frac{\mathrm{W}}{\mathrm{Q}}$ [Unit of emf : volt]

8.0Flow of charge through a conductor

(1) When there is no potential difference across a conductor, the directions of motion of free electrons present in it are randomly oriented such that there is no net drift of electrons in a particular direction. Hence, no electric current flows through it. [see figure (a)]

9.0Measuring potential difference

The device used to measure the potential difference across a current-carrying device (or conductor) is called voltmeter. Voltmeter is connected in parallel to the device across which potential difference is to be measured. (see figure)

Measuring potential difference across a resistor using a voltmeter.

10.0Measuring current

The device used to measure the electric current through device (or conductor) is called ammeter. Ammeter is connected in series with the device through which electric current is to be measured. (see figure)

• The voltmeter has a very large resistance, so that it draws very little current from the circuit and hence has very little effect on the voltage being measured. For an ideal voltmeter, resistance, $\mathrm{r}=\infty$.
• The ammeter has a very small resistance, so that it has very little effect on the current being measured. For an ideal ammeter, resistance, $r=0$.

Always take care that the positive terminal and the negative terminal of the voltmeter or an ammeter is always connected with the positive terminal and the negative terminal of the battery respectively.

11.0Electric circuits

A continuous path which consists of various electric devices like bulb, tube light, resistors, etc. connected with each other through conducting wires to the terminals of a source of emf like battery is called electric circuit.

Circuit elements

The various devices like electric bulbs, resistors, etc. connected in an electric circuit are called circuit elements. (see figure)

Circuit diagram

A diagram which indicates how different devices are connected in a circuit by using proper electric symbols for the devices is called a circuit diagram.

Building Concepts 2 When a conductor is joined with a battery, an electric current is setup in it due to flow of electrons through it. Does the battery supply electrons? Explain. Explanation: A battery does not supply electrons to the circuit. It establishes the electric field that exerts a force on electrons already in the wires and elements of the circuit that causes the flow of electrons.

Building Concepts 3 Identify the directions of current in each of the following (a) An electron moving clockwise (b) an alpha particle moving in north direction. Give reasons. Explanation: (a) Since electrons are negatively charged, the direction of electric current will be opposite to them. Thus, in the given situation, electric current will be anticlockwise. (b) Alpha particles are positively charged particles ( ${\mathrm{He}}^{2+}$ ions), thus, the direction of electric current will be same as that of alpha particles. In this situation, electric current will be in north direction.

• Some mathematical tools to be used in calculations. $\star {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}$ $\frac{a^m}{a^n}=a^{m-n}$ $\star {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-\mathrm{m}}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\mathrm{m}}$ $1\mathrm{m}=100\mathrm{cm};1\mathrm{cm}=10^{-2}\mathrm{m}$ $1\mathrm{m}=1000\mathrm{mm};1\mathrm{mm}=10^{-3}\mathrm{m}$ $1\mathrm{km}=1000\mathrm{m}$
• For a wire of cylindrical cross-section, Radius $=\frac{\text{ Diameter }}{2}$ Area of cross section of wire $=\pi r^2$
  
  Dimensions of a wire.

Numerical Ability 1

1. If a glass rod is charged to $+4.8\times 10^{-19}\mathrm{C}$, then n number of electrons get removed from it. Calculate value of $n$. Solution: Decode the problem Identify the terms $\mathrm{Q}=✓,\mathrm{e}=✓,\mathrm{n}=$ ? Apply the formula, $\mathrm{Q}=\mathrm{ne}$ Given $\mathrm{Q}=4.8\times 10^{-19}\mathrm{C};\mathrm{e}=1.6\times 10^{-19}\mathrm{C};\mathrm{n}=$ ? $\mathrm{Q}=\mathrm{ne}$ $\mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}=\frac{4.8\times 10^{-19}\mathrm{C}}{1.6\times 10^{-19}\mathrm{C}}=3$
2. How much work is done in moving a charge of 3 C across two points having a potential difference 18 V ? Solution: Decode the problem Identify the terms $\mathrm{Q}=✓,\mathrm{V}=✓,\mathrm{W}=$ ? Apply the formula $\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}$ $\times \mathrm{V}=3\times 18=54\mathbf{J}$ Given, charge, $\mathrm{Q}=3\mathrm{C}$; potential difference $=18\mathrm{V}$; work done, $\mathrm{W}=$ ? The amount of work done W in moving the charge is $\mathrm{W}=\mathrm{q}$
3. Atmospheric electricity : A lightning strike transfers 10 C of charge to the ground in 2 ms . Calculate the current during this lightning strike. Solution: Decode the problem Identify the terms $\mathrm{Q}=✓,\mathrm{t}=✓\mathrm{I}=$ ? Apply the formula, $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}$ Given, Charge, $\mathrm{Q}=10\mathrm{C}$; time, $\mathrm{t}=2\mathrm{ms}=2\times 10^{-3}\mathrm{s}$; current, I =? Now, current, $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{T}}=\frac{10}{2\times 10^{-3}}=5\times {10}^3\mathbf{A}$
4. If the current in a household appliance is 5 A , calculate the amount of charge that passes through the appliance in 1 hour. Solution: Decode the problem Identify the terms $\mathrm{I}=✓,\mathrm{t}=✓,\mathrm{Q}=$ ? Apply the formula, $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}$ Given, current, $\mathrm{I}=5\mathrm{A}$; time, $\mathrm{t}=1\mathrm{h}=3600\mathrm{s}$; charge. $\mathrm{Q}=$ ? Now, charge, $\mathrm{Q}=\mathrm{I}\times \mathrm{t}=5\times 3600=18000\mathrm{C}$ $=1.8\times 10^4\mathrm{C}$

12.0Electric resistance

The property of a material which offers opposition to the electric current and dissipates energy is called its 'electric resistance'.

T The ratio of applied potential difference (voltage) to the resulting electric current in a closed circuit is called 'electric resistance'.

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

Resistance is the property of a conductor to resist the flow of charge through it.

Unit of resistance SI unit : Ohm ( $\Omega$ ) $1\mathrm{ohm}=1\mathrm{volt}/$ ampere or $1\Omega =1{\mathrm{VA}}^{-1}$

Definition of 1 \mathbf{~ o h m} When the potential difference of 1 volt is applied across the ends of a conductor and the current produced in it is 1 ampere, then the resistance of the conductor is 1 ohm .

13.0Ohm's law

(given by German physicist George Simon Ohm in 1827) According to Ohm's law, 'the current through certain conductors is directly proportional to the potential difference between its ends at a constant temperature'. [see figure (a) and (b)] $I\propto V$ or $V\propto I$ or $V=IR$ or $R=\frac{V}{I}$

Where, R is constant called resistance of the conductor.

Ohm's law is not a fundamental law, it is only an empirical law that holds approximately good for many substances particularly metals. (see figure)

• The resistors which obey Ohm's law are called linear resistors. The resistors which do not obey Ohm's law are called non-linear resistors.

Slope of graph $=\tan \theta =\frac{\text{ Perpendicular }}{\text{ base }}=\frac{\text{ Change in }y\text{-axis }}{\text{ Change in }x\text{-axis }}=\frac{\text{ Rise }}{\text{ Run }}$

The vertical change between two points is called the rise, and the horizontal change is called the run. Slope of $AB=\frac{BC}{AC}=\frac{y_2-y_1}{x_2-x_1}$

• More the slope of the V-I graph, more is the resistance of the conductor shown in the graph and vice-versa.
• Always study the graph between $V$ and I properly. If $V$ is on $y$ axis and $I$ is on $x$-axis then $R=\frac{V}{I}=\tan \theta$. But if $I$ is on $y$ axis and $V$ is on $x$ axis then $R=\frac{1}{\tan \theta }$. Take care of the units of voltage and current while calculating resistance. $1\mathrm{mV}=1\times 10^{-3}\mathrm{V};1\mu \mathrm{A}=10^{-6}\mathrm{A},1\mathrm{mA}=10^{-3}\mathrm{A}$.

Numerical Ability 2

1. The graph shown below is V-I graph inclined to $x$-axis. What is the resistance of the graph?
   
   Solution: Decode the problem Identify the terms Perpendicular is BC, base is AC. Apply the formula, $\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$ $R_{AB}=$ Slope of line $AB=\frac{\text{ Perpendicular }}{\text{ base }}=\frac{\text{ Change in }y\text{-axis }}{\text{ Change in }x\text{-axis }}$ $\frac{{\mathrm{V}}_2-{\mathrm{V}}_1}{{\mathrm{I}}_2-{\mathrm{I}}_1}=\frac{20-0}{10-0}=\frac{20}{10}=2\Omega$

14.0Resistors

A physical device which has the principal characteristic of offering electric resistance is called 'resistor'.

Materials used for resistors

(1) Alloys like manganin, constantan, nichrome, etc.; used in wire wound resistors. (2) Carbon resistors ; compact and low cost. (3) Aluminium or copper wires; low resistance conductors used to make connecting wires/electrical transmission lines.

• Resistance ' $R$ ' is not a material property that is, its value changes from sample to sample for a given material. Resistance depends on the nature of substance, its shape and size (geometrical factors like length, cross-sectional area).

Resistance is directly proportional to the length ( $\ell$ ) of a conductor and inversely proportional to its area of cross-section (A). $\mathrm{R}\propto \ell ,\mathrm{R}\propto \frac{1}{\mathrm{A}}$ or $\mathrm{R}\propto \frac{\ell }{\mathrm{A}}$ or $\mathbf{R}=\mathbf{\rho }\frac{\ell }{\mathbf{A}}$

Where, $\rho$ is a constant called resistivity of material. Resistance ' R ' depends on : (1) length of conductor (2) cross-sectional area (3) type of material (4) temperature

15.0Resistivity (specific resistance)

It is a characteristic property of a material rather than that of a particular specimen of a material. It depends on physical conditions such as temperature and pressure.

If $\ell$ is 1 unit length and $A$ is 1 square unit area, then, $R=\rho$ Resistivity of a conductor is 'the resistance offered by a uniform conducting wire having unit length and unit area of cross-section'.

Unit of resistivity SI unit : ohm-meter ( $\Omega \mathrm{m}$ )

• Resistance and resistivity of substances depend on temperature. For metals, they increase with the increase in temperature. For insulators or semiconductors, they decrease with the increase in temperature. $\star$ Resistivity does not depend on the physical dimensions or shape of the material. If we stretch or fold the wire, it's resistivity does not change.

Active Physics 1

1. Let us setup a circuit consisting of a resistance R; an ammeter A connected in series with R; a voltmeter connected in parallel with $R$; a number of cells each having potential difference V that are connected in series. [see figure (a)]
2. First we use only one cell in the circuit and note down the value of current in the circuit and potential difference across resistance R. Repeat this experiment with two cells, three cells, four cells, etc. Note down the successive readings in the voltmeter and ammeter. You will find that for one cell in the circuit, if current in the circuit is I and potential difference is V , then for two, three, four cells, their values will be $2\mathrm{I},3\mathrm{I},4\mathrm{I}$ and $2\mathrm{V},3\mathrm{V},4\mathrm{V}$ respectively.
3. Now, plot a graph between current and voltage. You will get a straight line, which verifies the Ohm's law. [see figure (b)]
   

Active Physics 2

1. Connect a circuit which consists of a cell, an ammeter and a wire of high resistance (say, nichrome) having length $L$ and a switch (see figure).
   
2. Now, 'switch on' and note down the value of current in the ammeter. Replace the wire by another one having same cross-sectional area but length ' 2 L '. You will find that current has now decreased to one half. This means resistance is increased in the circuit, i.e., R is directly proportional to length. $\mathrm{R}\propto \mathrm{L}$
3. Now, take a nichrome wire of larger cross-sectional area but having same length L. You will find that current in the circuit has been increased. This means resistance is decreased in the circuit, i.e. R is inversely proportional to cross-sectional area. $\mathrm{R}\propto \frac{1}{\mathrm{A}}$
4. Take another wire instead of nichrome having same cross-sectional area and length as that of nichrome. You will find that the current in the ammeter has changed (may get decreased or increased). This means resistance depends on the nature of material of the wire.

16.0Effect of stretching of a wire on resistance

Let a wire of length ${\ell }_1$ and cross-sectional area ${\mathrm{A}}_1$ be stretched to a length ${\ell }_2$ and its crosssectional area becomes ${\mathrm{A}}_2$ such that ${\ell }_2=\mathrm{n}{\ell }_1$ (see figure).

Now, volume after stretching = volume before stretching or ${\ell }_2{\mathrm{A}}_2={\ell }_1{\mathrm{A}}_1$ or $\left(\mathrm{n}{\ell }_1\right){\mathrm{A}}_2={\ell }_1{\mathrm{A}}_1$ or $A_2=\frac{A_1}{n}$ Initial resistance, ${\mathrm{R}}_1=\rho \frac{{\ell }_1}{{\mathrm{A}}_1}$ Final resistance, ${\mathrm{R}}_2=\rho \frac{{\ell }_2}{{\mathrm{A}}_2}=\rho \frac{\mathrm{n}{\ell }_1}{\left({\mathrm{A}}_1/\mathrm{n}\right)}={\mathrm{n}}^2\left(\rho \frac{{\ell }_1}{{\mathrm{A}}_1}\right)$ [Using (1) and (2)] $\text{ or }{\mathbf{R}}_2={\mathbf{n}}^2{\mathbf{R}}_1$

• For a given wire, if the length is stretched to n times then the area is reduced to $\frac{1}{\mathrm{n}}$ times. For example, for a given wire of length ' $\ell$ ' and area of cross section ' $A$ ', if the wire is stretched so that its length becomes 5 times, then the area of cross-section of the wire will be $\frac{A}{5}$. $\star$ If wire is folded such that its length becomes $\frac{\ell }{n}(1/n$ of the original length $)$, then the area of cross-section of the wire will be nA.

For example if the length of wire on folding becomes $\frac{1}{3^{\text{rd }}}$ of the original length, then area of cross-section will be 3A.

Numerical Ability 3

1. The potential difference between the terminals of an electric heater is 80 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V ? Solution: Decode the Problem Case -I Identify the terms $\mathrm{V}=✓,\mathrm{I}=✓,\mathrm{R}=$ ? Apply the formula, $\mathrm{V}=\mathrm{IR}$ Case -II Identify the terms $\mathrm{V}=✓,\mathrm{R}=✓,\mathrm{I}=$ ? Apply the formula $\mathrm{V}=\mathrm{IR}$ Given, potential difference, $\mathrm{V}=80\mathrm{V}$; current, $\mathrm{I}=4\mathrm{A}$. According to Ohm's law, $R=\frac{V}{I}=\frac{80}{4}=20\Omega$ Now, the new potential difference, ${\mathrm{V}}^{\prime }=120\mathrm{V}$ New current, ${\mathrm{I}}^{\prime }=\frac{{\mathrm{V}}^{\prime }}{\mathrm{R}}=\frac{120}{20}$ $=6\mathrm{A}$
2. Resistance of a metal wire of length 220 m is $7\Omega$ at $20^{\circ }\mathrm{C}$. If the diameter of the wire is 0.8 mm , what will be the resistivity of the metal at that temperature? Solution: Decode the problem Identify the terms $R=✓,D=✓$ (radius can be calculated through the diameter given), $\ell =✓$, $A=✓\left(A=\pi r^2\right),\delta =?$Apply the formula, $R=\rho \frac{\ell }{A}$ Given, resistance, $\mathrm{R}=7\Omega$; diameter, $\mathrm{d}=0.8\mathrm{mm}$; length, $\ell =220\mathrm{m}$. Now, radius of the wire, $r=\frac{\text{ diameter }}{2}=\frac{0.8}{2}=0.4\mathrm{mm}$ $=0.4\times 10^{-3}\mathrm{m}$ Resistivity of a metallic wire is given by, $$\begin{gathered} \rho =\frac{\mathrm{RA}}{\ell }=\frac{\mathrm{R}\left(\pi {\mathrm{r}}^2\right)}{\ell }=\frac{7\times (22/7)\times {\left(0.4\times 10^{-3}\right)}^2}{220} \\ =\frac{7\times 22\times {\left(4\times 10^{-1}\times 10^{-3}\right)}^2}{7\times 22\times 10} \\ =\frac{{\left(4\times 10^{-4}\right)}^2}{10}=\frac{16\times 10^{-8}}{10} \\ =1.6\times 1{10}^{-8}\mathbf{\Omega }\mathbf{m} \end{gathered}$$
3. An $8\Omega$ resistance wire is doubled on itself. Calculate the value of the new resistance offered by the wire. Solution: Decode the problem Identify the terms $\ell =✓,A=✓,R=$ ? Apply the formula, $R=\rho \frac{\ell }{A}$ ${\mathrm{R}}^{\prime }=\rho \frac{{\ell }^{\prime }}{{\mathrm{A}}^{\prime }}=\rho \frac{(\ell /2)}{2\mathrm{A}}=\frac{1}{4}\left(\rho \frac{\ell }{\mathrm{A}}\right)=\frac{1}{4}\mathrm{R}=\frac{1}{4}\times 8=2\mathbf{\Omega }$ Let the length and area of cross-section of the original wire be $\ell$ and A respectively (see figure).
   
   (a) Original wire
   
   (b) Wire is doubled on itself Original resistance, $R=\rho \frac{\ell }{A}$ When a wire is doubled on itself, its length becomes halved and area of cross-section becomes doubled i.e., the new length, ${\ell }^{\prime }=\frac{\ell }{2}$ and new area of cross-section, ${\mathrm{A}}^{\prime }=\mathrm{A}+\mathrm{A}=2\mathrm{A}$. New resistance,
4. A $3\Omega$ resistance wire is stretched to 3 times. Calculate the value of the new resistance offered by the wire. Solution: Decode the problem Identify the terms $\ell =✓,A=✓,R=?$ Apply the formula, $\mathrm{R}=\rho \frac{\ell }{\mathrm{A}}$ Let the length and area of cross-section of the original wire be $\ell$ and A respectively (see figure). Original resistance, $R=\rho \frac{\ell }{A}$ When a wire is stretched 2 times, its length becomes ${\ell }^{\prime }=3\ell$ and area of cross-section becomes ${\mathrm{A}}^{\prime }=\frac{\mathrm{A}}{3}$ New resistance, ${\mathrm{R}}^{\prime }=\rho \frac{{\ell }^{\prime }}{{\mathrm{A}}^{\prime }}=\rho \frac{(3\ell )}{\frac{\mathrm{A}}{3}}=9\left(\rho \frac{\ell }{\mathrm{A}}\right)=9\mathrm{R}=9\times 3=27\mathbf{\Omega }$

Building Concepts 4 Two materials have different resistivities. Two wires of the same length are made, one from each of the materials. Is it possible for each wire to have the same resistance? Explanation: Two wires of the same length are made, one from each of the materials. The resistance of a wire is given by, $R=\rho \frac{\ell }{\mathrm{A}}$, where $\rho$ is the resistivity of the wire material, and $\ell$ and A are respectively, the length and cross-sectional area of the wire. Even when the wires have the same length, they may have the same resistance, if the cross-sectional areas of the wires are chosen so that the ratio $\rho \frac{\ell }{\mathrm{A}}$ is the same for each. That is, ${\rho }_1\frac{\ell }{A_1}={\rho }_2\frac{\ell }{A_2}\text{ or }\frac{A_2}{A_1}=\frac{{\rho }_2}{{\rho }_1}$ This is the condition for each wire of different material to have the same resistance when they have same length. Two wires made of iron of different sizes may have different resistances but they have same resistivities. This is because resistance differ from specimen to specimen while resistivity is constant for a given material at a given temperature.

17.0Resistors in series

Let us consider three resistors having resistances ${\mathrm{R}}_1,{\mathrm{R}}_2$ and ${\mathrm{R}}_3$ respectively joined in series. Let $V_1,V_2,V_3$ are voltages across resistors $R_1,R_2,R_3$ respectively. The current through resistors $R_1,R_2$ and $R_3$ are $I_1,I_2$ and $I_3$ respectively. Since the resistors are joined in series, thus, current through each resistor is same because current entering at one point (end) is equal to current leaving at the other point (end). (see figure)

A series combination of resistors ${\mathrm{I}}_1={\mathrm{I}}_2={\mathrm{I}}_3=\mathrm{I}$ (let) Potential difference ' $V$ ' between the point $A$ and point $B$ is the sum of the voltages across $R_1$, ${\mathrm{R}}_2$ and ${\mathrm{R}}_3$. $\mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2+{\mathrm{V}}_3$ Let ' $R$ ' be the equivalent resistance of the whole combination. Equivalent resistance: The resistance of a single resistor that can replace a combination of resistors in any given circuit without any change in potential difference across the terminals and current through the circuit is called 'equivalent resistance'. $\therefore \mathrm{V}={\mathrm{IR}}_{\mathrm{s}}$ From (2) and (3) we get, $$\begin{gathered} I_s=V_1+V_2+V_3 \\ \text{ or }{\mathrm{IR}}_s=I{R}_1+I{R}_2+I{R}_3[V=IR] \\ \text{ or }I{R}_s=I\left(R_1+R_2+R_3\right) \\ \text{ or }{R}_s=R_1+R_2+R_3 \end{gathered}$$ General formula for ' $n$ ' resistors in series : $R_s=R_1+R_2+R_3+\dots +R_n$

Active Physics 3

1. Connect three resistors $\left(R_1,R_2,R_3\right)$ in series. Connect this combination with a battery, a plug key K and an ammeter A (see figure).
2. Now, connect a voltmeter across the series combination of resistors between $X$ and $Y$ points [see figure (a)]. Plug the key and note down the reading in the voltmeter. This reading (V) is the total potential difference across the combination.
3. Now, take out the plug key and disconnect the voltmeter. Insert the voltmeter across the ends X and P of the first resistor [see figure (b)]. Plug the key and measure the potential difference ( ${\mathrm{V}}_1$ ) across the first resistor. Similarly, measure the potential differences ( $V_2$ and $V_3$ ) across the other two resistors, one by one.
4. You will find that the readings $V_1,V_2$ and $V_3$ are different. Now, add $V_1,V_2$ and $V_3$. You will find that sum of these potential differences is equal to the total potential difference (V) across the combination. That is, $V=V_1+V_2+V_3$
5. Also, note the ammeter reading [see figure (a)]. Now, change the position of ammeter to anywhere in between the resistors [see figure (c)]. Note the ammeter reading each time. You will find that, the reading of ammeter remains the same i.e., the current flowing through each resistor is same in a series circuit and it is equal to the total current flowing in the circuit.
   
   (a)
   
   (b)
   
   (c)

Important points related to series combination

1. Current is same in every part of circuit. (a) That is, ${\mathrm{I}}_1={\mathrm{I}}_2={\mathrm{I}}_3=$ $=\mathrm{In}$ (b) Also, $\frac{V}{R}=I$ constant or $V\propto R$ (c) ${\mathrm{V}}_1:{\mathrm{V}}_2:{\mathrm{V}}_3={\mathrm{R}}_1:{\mathrm{R}}_2:{\mathrm{R}}_3$
2. Equivalent resistance is equal to sum of the individual resistances.
3. Total potential difference across the combination is equal to the sum of potential difference across each resistor.

That is, $V=V_1+V_2+V_3+\dots ..+V_n$ 4. In series combination, if one resistance gets 'open', the current in the whole circuit will be zero as circuit breaks. $\infty$ If there are ' $n$ ' equal resistors ( R ) connected in series, then, their equivalent resistance is ( n R ). If V is the potential difference applied across the combination, then potential difference across each resistor is $\left(\frac{V}{n}\right)$.

18.0Resistors in parallel

Let us consider three resistors having resistances ${\mathrm{R}}_1,{\mathrm{R}}_2,{\mathrm{R}}_3$ respectively. Let the voltage across the combination is ' $V$ ' and currents through $R_1,R_2,R_3$ are $I_1,I_2,I_3$. Voltage across all the resistors is same as all of them have same terminal points (A and B). (see figure)

A parallel combination of resistors. ${\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_3=\mathrm{V}$ (let) The total current ' $I$ ' entering through ' $A$ ' is divided among the three resistors ( ${\mathrm{I}}_1,{\mathrm{I}}_2$ and ${\mathrm{I}}_3$ ). Thus, the total current ' $I$ ' is the sum of individual currents through $R_1,R_2$ and $R_3$. $\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$ Let the equivalent resistance of the whole combination be ${\mathrm{R}}_{\mathrm{p}}$. $\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{p}}}$ From (2) and (3), we get, $\frac{V}{R_p}=I_1+I_2+I_3$ or $\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{p}}}=\frac{\mathrm{V}}{{\mathrm{R}}_1}+\frac{\mathrm{V}}{{\mathrm{R}}_2}+\frac{\mathrm{V}}{{\mathrm{R}}_3}\left[\because \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\right]$ or $\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{p}}}=\mathrm{V}\left(\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}\right)$ or $\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}$ General formula for n resistors in parallel : $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots +\frac{1}{R_n}$

Important points related to parallel combination

1. The sum of the reciprocals of the separate resistances is equal to the reciprocal of equivalent resistance.
2. The voltage across each resistor of a parallel combination is the same, and is equal to the voltage across the whole group. (a) ${\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_3=\dots ....={\mathrm{V}}_{\mathrm{n}}=$ constant (b) IR $=\mathrm{V}=\mathrm{constant}$ or $\mathrm{I}\propto \frac{1}{\mathrm{R}}$
3. In parallel combination, ${\mathrm{I}}_1:{\mathrm{I}}_2:{\mathrm{I}}_3=\frac{1}{{\mathrm{R}}_1}:\frac{1}{{\mathrm{R}}_2}:\frac{1}{{\mathrm{R}}_3}$
4. The total current is the sum of currents flowing in different branches. $\text{ That is, }\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3+\dots ....{\mathrm{In}}_n$
5. For two resistors $R_1$ and $R_2$ in parallel, their equivalent resistance is given by, $R_e=\frac{R_1{R}_2}{R_1+R_2}$
6. In case of resistors in parallel, if one resistance becomes 'open', all others will work as usual. If ' $n$ ' equal resistors ( $R$ ) are connected in parallel, then, their equivalent resistance is $\left(\frac{R}{n}\right)$. If I is the total current passing through the combination, then the current through each resistor is $\left(\frac{I}{n}\right)$.

• We have seen that in a series circuit the current is constant throughout the electric circuit. Thus it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly. Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works.
• On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has different resistance and requires different current to operate properly.

Active Physics 4

1. Make a parallel combination, $XY$, of three resistors having resistors $\left(R_1,R_2,R_3\right)$. Connect it with a battery, a plug key and an ammeter [Figure (a)]. Also, connect a voltmeter in parallel with the combination of resistors.
2. Now, plug the key and note the ammeter reading (I), this reading is the total current flowing in the circuit. Also, take the voltmeter reading (V), this reading is the potential difference across the parallel combination.
3. Now, take out the key from the plug and remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor ${\mathrm{R}}_1$ and a voltmeter in parallel with ${\mathrm{R}}_1$ [Figure (b)]. The potential difference across resistor ${\mathrm{R}}_1$ is also V. Similarly, we will find that potential difference across resistor ${\mathrm{R}}_2$ and ${\mathrm{R}}_3$ is also V by connecting voltmeter across them. Thus, in parallel combination, the potential difference across each resistor is same as the total potential difference V .
4. Now, note the ammeter reading ( ${\mathrm{I}}_1$ ), this reading is the current passing through ${\mathrm{R}}_1$. Similarly, measure the currents through ${\mathrm{R}}_2$ and ${\mathrm{R}}_3$. Let these currents be ${\mathrm{I}}_2$ and ${\mathrm{I}}_3$, respectively. You will find that sum of these currents is equal to total current (I) flowing in the circuit. That is, $\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$.
   
   (a)
   
   (b)

Numerical Ability 4

1. In the circuit shown (figure), find (a) equivalent resistance of the network (b) total current flowing in the circuit (c) current flowing in $6\Omega$ resistor (d) potential difference across $8\Omega$ resistor.
   
   Solution: Decode the problem (a) Identify the types of combination Apply the formula, ${\mathrm{R}}_{\mathrm{s}}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3$ $\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}$ (b) Identify the terms $\mathrm{V}=✓,\mathrm{R}=✓$ Apply the formula, $\mathrm{V}=\mathrm{IR}$ (c) Identify the terms $R=✓,I=✓$ Apply the formula, $\mathrm{V}=\mathrm{IR}$ (d) Identify the terms $R=✓,I=✓$ Apply the formula, $\mathrm{V}=\mathrm{IR}$ (a) $6\Omega$ and $12\Omega$ are joined in parallel, thus, their equivalent resistance is, ${\mathrm{R}}_1=\frac{6\times 12}{6+12}=\frac{6\times 12}{18}=4\mathbf{\Omega }$ Now, ${\mathrm{R}}_1$ and $8\Omega$ resistors are in series [see figure (b)], thus, their equivalent resistance is, ${\mathrm{R}}_{\mathrm{e}}={\mathrm{R}}_1+8=4+8=12\mathbf{\Omega }$ (b) Total current in the circuit, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{1.5}{12}=0.125\mathrm{A}$ (c) Potential difference across $6\Omega$ (or $12\Omega$ ) resistor, $\mathrm{V}=\mathrm{I}\times {\mathrm{R}}_1=0.125\times 4=0.5\mathrm{V}$ Current through $6\Omega$ resistor, $I=\frac{V}{R}=\frac{0.5}{6}=0.083\mathrm{A}$ (d) Potential difference across $8\Omega$ resistor, $\mathrm{V}=\mathrm{I}\times 8=0.125\times 8=1V$

• To get maximum resistance, resistors must be connected in series and to get minimum resistance, resistors must be connected in parallel.

Building Concepts 5 Figure shows a combination of four identical bulbs joined with a battery. Compare the brightness of the bulbs shown. What happens if bulb A fails, so that it cannot conduct current? What happens if bulb C fails? What happens if bulb D fails?

• If positive and negative terminals of a battery are joined by a connecting wire i.e., it is shortcircuited, then large current will flow through it as resistance of the circuit becomes negligible.
• If both the terminals of a device (like a bulb) are joined by a connecting wire i.e., they are joined directly then, the device is said to be short circuited and potential difference across it becomes zero.

19.0Heating effect of electric current

We know that to maintain the current, the source (cell or battery) has to keep expending its energy. A part of the source energy is consumed into useful work like in rotating the blades of an electric fan, producing light and sound in television, etc. Rest of the source energy is converted to heat to raise the temperature of the device. (G) If an electric circuit is purely resistive, that is, it consists of resistors only connected to a battery, the source energy gets dissipated entirely in the form of heat. This effect is utilised in devices such as electric geysers, electric heater, electric iron, etc. (o) The conversion of a part (or whole) of the electric energy into heat energy when electric current flows through a device is called heating effect of current. Ge Resistivities of alloys are generally higher than that of their constituent metals, they produce more heat as compared to a pure metal. Also, they do not burn (oxidise) readily at high temperatures. Thus, they are used in electrical heating devices like geyser, electric iron, toasters, etc.

Electric power

The work done per unit time by a source of emf (like a battery) in order to maintain electric current in a circuit is called 'electric power'.

Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V (figure). Let t

Heating effect of current be the time during which a charge Q flows across. The work done in moving the charge $Q$ through a potential difference $V$ is given by, $\mathrm{W}=\mathrm{Q}\times \mathrm{V}\dots$ (1) $\left(\because \mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}\right)$ Thus, the power input to the circuit by the source is $\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{Q}\times \mathrm{V}}{\mathrm{t}}=\left(\frac{\mathrm{Q}}{\mathrm{t}}\right)\times \mathrm{V}=\mathrm{I}\times \mathrm{V}$ or $\mathrm{P}=\mathrm{VI}$

Unit of electric power SI unit : Watt Here, 1 Watt = 1 volt-ampere = 1 VA 1 kilowatt = 1000 watt

• If the electric current through a device is one ampere and the potential difference across it is one volt, then its electric power is one Watt. Also, power consumed by a resistor can be written as, $\mathrm{P}=\mathrm{VI}=\frac{{\mathrm{V}}^2}{\mathrm{R}}={\mathrm{I}}^2\mathrm{R}$ [Using, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ and $\mathrm{V}=\mathrm{IR}$ ]
• The maximum current that can flow through a device is called the current capacity of that device. If an electric bulb is rated ' $V$ ' volts and ' P ' watts, then its current capacity $(\mathrm{Ic})$ is given by, ${\mathrm{IC}}_{\mathrm{C}}=\mathrm{P}/\mathrm{V}$.
• Every electric device is mentioned with a specified power and a specified voltage ; it works properly only if it runs at the specified voltage.
• If applied voltage is greater than specified voltage for a electric device, then the actual current exceeds its current capacity, the device will get damaged. If applied voltage is less than the specified voltage, then the current in the circuit will be less than the required; the device will not work properly.

Electric energy

The total energy supplied by a source of emf in order to maintain the electric current in the circuit in a given time is called 'electric energy'. $\mathrm{E}=\mathrm{P}\times \mathrm{t}=\mathrm{V}\times \mathrm{I}\times \mathrm{t}$ Unit of electric energy SI unit : Joule Here, 1 Joule $=1$ volt-ampere-sec $=1\mathrm{VAs}$ Other unit : 1 Watt-hour $=3600\mathrm{J}$ 1 Kilo Watt Hour $(\mathrm{KWh})=1000$ watt $\times 3600\sec =3.6\times 10^6\mathrm{J}$

• Long distance power transmission is done at high voltages because at high voltages, current flowing through the transmission wires is less. As a result, heat losses ( $\mathrm{H}\propto {\mathrm{I}}^2$ ) will be less.
• Kilo Watt Hour (KWh) is the commercial unit of electric energy. If an electric device of 1 kW is used for 1 hr , then the energy consumed is 1 KWh .

20.0Joule's law of heating

The energy supplied to the circuit by a source (like a battery) in time $t$ is given by,

$\mathrm{E}=\mathrm{P}\times \mathrm{t}=\mathrm{V}\times \mathrm{I}\times \mathrm{t}[\because \mathrm{P}=\mathrm{V}\times \mathrm{I}]$

This energy (E) gets dissipated in the resistor as heat. Thus, for a steady current I, the amount of heat H produced in time t is $\mathrm{H}=\mathrm{V}\times \mathrm{I}\times \mathrm{t}$ Applying Ohm's law, i.e., $\mathrm{V}=\mathrm{I}\times \mathrm{R}$, we get, $\mathbf{H}={\mathbf{I}}^2\mathbf{Rt}$ This is known as Joule's law of heating. © According to Joule's law of heating, the heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor.

21.0Applications of thermal effects of current

Electric bulb

The electric heating or joules heating is used in producing light in 'electric bulb'. Here, the filament must retain as much of the heat generated as possible, so that it gets very hot and emits light. It must not melt at such high temperature. Thus, an electric bulb consists of a filament of a strong metal with high melting point such as 'tungsten' (melting point $3380^{\circ }\mathrm{C}$ ) sealed in a glass bulb. The bulbs are usually filled with chemically inactive 'nitrogen' or 'argon' gases which prevents the oxidation of filament at high temperatures, thereby, increasing the life of bulb. Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated.

If an electric bulb is rated ' $V$ ' volts and ' P ' watts, then its resistance is given by $\mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}$. For example, for a $40\mathrm{W},220\mathrm{V}$ bulb, $\mathrm{R}=\frac{(220{)}^2}{40}=1210\Omega$.

Electric fuse and its action

It is a safety device used to prevent the electric appliances against excessive electric currents. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. Usually, a metallic conducting wire (fuse wire) made of Tin (25%) and Lead (75%) having low melting point is used. The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends. It is put in series with the electric device in the circuit. If a current larger than the specified value of current capacity flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit and thus, the electric device in the circuit is prevented from getting damaged.

• The fuses used for domestic purposes are rated as $1\mathrm{A},2\mathrm{A},3\mathrm{A},5\mathrm{A},10\mathrm{A}$, etc. © The current capacity of a fuse is independent of its length, it is proportional to its radius. More the radius, more will be the current capacity and vice-versa. © For an electric iron which consumes 1 kW electric power when operated at 220 V , a current of $\left(\frac{1000}{220}\right)\mathrm{A}$, that is, 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.
• We may think that electrons are consumed in an electric circuit. This is wrong! We pay the electricity board or the electric company to provide energy to move electrons through the electric devices like electric bulb, fan, televisions, refrigerators, etc. We pay for the energy that we consume in the electric devices.

Numerical Ability 5

1. An electric iron consumes energy at a rate of 960 W when heating is at the maximum rate and 480 W when the heating is at the minimum. The voltage is $220\mathrm{V}$. What are the current and the resistance in each case? Solution: Decode the problem Identify the given terms, $P=✓,V=✓$ Apply the formula, $\mathrm{P}=\mathrm{VI}$ $\mathrm{V}=\mathrm{IR}$ We know that the power input is $\mathrm{P}=\mathrm{VI}\therefore \mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ (a) When heating is at the maximum rate, Current, $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{960}{220}=4.36\mathrm{A}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{220}{4.36}=50.46\Omega$ (b) When heating is at the minimum rate, Current, $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{480}{220}=2.18\mathrm{A}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{220}{2.18}=100.9\mathbf{\Omega }$
2. 200 J of heat is produced every second in a $8\Omega$ resistance. Find the potential difference across the resistor. Solution: Decode the problem Identify the given terms, $H=✓,R=✓,t=✓$ Apply the formula, $\mathrm{H}={\mathrm{I}}^2\times \mathrm{Rt}$ Now, potential difference, $\mathrm{V}=\mathrm{I}\times \mathrm{R}=5\times 8=40\mathrm{V}$ Given, heat, $\mathrm{H}=200\mathrm{J}$; resistance, $\mathrm{R}=8\Omega$; time, $\mathrm{t}=1\mathrm{s}$. Now, $\mathrm{H}={\mathrm{I}}^2\mathrm{R}$ t or ${\mathrm{I}}^2=\frac{\mathrm{H}}{\mathrm{Rt}}=\frac{200}{8\times 1}=25$ or $\mathrm{I}=\sqrt{25}=5\mathrm{A}$
3. Some devices used at a home are given along with their numbers, power ratings and usage time.

Find the total units (in KWH) of energy consumed for 30 days. What is the cost of the total energy consumed for $30$ days if one unit costs Rs 3.00? Solution: Decode the problem Identify the given terms, $P=✓,V=✓$ Apply the formula, $\mathrm{E}=\mathrm{P}\times \mathrm{t}$ First we must convert power in KW for each device and then find the power consumed by them per day. Device energy consumed/day in KWH Refrigerator $1\times \frac{400}{1000}\mathrm{KW}\times 8\mathrm{hr}=3.2$ Electric bulb $2\times \frac{40}{1000}\mathrm{KW}\times 4\mathrm{hr}=3.2$ Tube light $4\times \frac{60}{1000}\mathrm{KW}\times 4\mathrm{hr}=0.96$ Fan $4\times \frac{100}{1000}\mathrm{KW}\times 6\mathrm{hr}=2.4$ Total energy consumed/day by all devices $=6.88\mathbf{KWH}$ Total energy consumed by all devices in 30 days $=6.88\times 30=206.4\mathbf{KWH}$ Cost of total energy consumed for 30 days $=$ Rs. $3\times 206.4=$ Rs. 619.20

4. An electric bulb is connected to a 220 V generator. The current is 0.50 A . What is the power of the bulb? What is its resistance? Solution: Decode the problem Identify the given terms, $V=✓,I=✓$ Apply the formula, $\mathrm{P}=\mathrm{VI}$ $\mathrm{V}=\mathrm{IR}$ We know that power, $\mathrm{P}=\mathrm{V}\times \mathrm{I}$ or $\mathrm{P}=220\mathrm{V}\times 0.50\mathrm{A}$ $=110\mathrm{W}$ Now, resistance, $\mathrm{R}=\frac{\mathrm{V}}{1}=\frac{220}{0.5}=440\Omega$
5. How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts. Solution: Decode the problem (i) Identify the given terms, $\mathrm{V}=✓,\mathrm{R}=✓,\mathrm{I}=$ ? Apply the formula, $\mathrm{V}=\mathrm{IR}$ (ii)Identify the given terms, $\mathrm{V}=✓,\mathrm{R}=✓,\mathrm{P}=$ ? Apply the formula, $\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}$ Given; $\mathrm{V}=220\mathrm{V},\mathrm{R}=55\Omega$ $\mathrm{V}=\mathrm{IR}\Rightarrow \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\mathrm{I}=\frac{220}{55}=4\mathrm{A}$ Wattage of electric iron = Power $\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{(220{)}^2}{55}=880\mathbf{W}$
6. Two identical resistors, each of resistance $5\Omega$ are connected in (i) series and (ii) parallel, in term to a battery of 10 V , calculate the ratio of the power consumed in the combination of resistors in each case. Solution: Decode the problem Identify the types of combination. Apply the formula, ${\mathrm{R}}_{\mathrm{s}}={\mathrm{R}}_1+{\mathrm{R}}_2$ $\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}$ Identify the terms, $R=✓,V=✓,P=$ ? Apply the formula, $\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}$ Given; ${\mathrm{R}}_1={\mathrm{R}}_2=5\Omega ,\mathrm{V}=10\mathrm{V}$ (i) When connected in series, ${\mathrm{R}}_{\mathrm{s}}=5\Omega +5\Omega =10\Omega$ ${\mathrm{P}}_{\mathrm{s}}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{s}}}=\frac{(10{)}^2}{10}=10\mathbf{W}$ (ii) When connected in parallel, $\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\Rightarrow {\mathrm{R}}_{\mathrm{p}}=\frac{5}{2}\Omega$ ${\mathrm{P}}_{\mathrm{p}}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{p}}}=\frac{(10{)}^2}{5/2}=40\mathrm{W}$ $\therefore$ The required ratio $=\frac{{\mathrm{P}}_{\mathrm{s}}}{{\mathrm{P}}_{\mathrm{p}}}=\frac{10}{40}=\frac{1}{4}$

22.0Concept Map

Important units, formulae, and quantity

23.0Some Basic Terms

1. Charge :- It is the physical property of the matter that causes it to experience a force, when placed in an electromagnetic field.
2. Quantization :- Quantization is the concept that a physical quantity can have only certain discrete values.
3. Electric Potential :- The amount of work energy needed per unit of electric charge to move the charge from a reference point to a specific point in an electric field.
4. Scalar :- A physical quantity that is completely described by its magnitude.
5. Viscosity :- Viscosity is a measure of a fluid's resistance to flow.
6. Voltage :- Voltage describes the "pressure" that pushes electrons.
7. Dissipates :- To break up and scatter or vanish
8. Semiconductors :- A semiconductor is a material that has an electrical conductivity value falling between that of a conductor, such as copper, and an insulator, such as glass.
9. Equivalent :- A like or equal in number, value, or meaning or having the same numerical value.
10. Split:- To divide or share something
11. Transmission :- The process of passing something from one person or place to another.
12. Steady :- Staying the same

24.0SOLVED EXAMPLES

1. What is the current flowing through a conductor if one million electrons are crossing in one millisecond through a cross-section of it? Solution: Given, no. of electrons, $\mathrm{n}=$ one million $=10^6$; charge on electron, $\mathrm{e}=1.6\times 10^{-19}\mathrm{C}$; time, $\mathrm{t}=$ one milli second $=10^{-3}\mathrm{s}$ Now, charge, $Q=$ ne and current, $I=\frac{Q}{t}$ $$\begin{gathered} \therefore \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{ne}}{\mathrm{t}}=\frac{10^6\times 1.6\times 10^{-19}}{10^{-3}} \\ =1.6\times {10}^{-10}\mathbf{A} \end{gathered}$$
2. A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0\times 10^{-7}{\mathrm{m}}^2$ and its resistance is $50\Omega$. Find out its resistivity. Solution: Given, length, $\mathrm{L}=15\mathrm{m}$; resistance, $R=50\Omega$; area of cross - section, $\mathrm{A}=6.0\times 10^{-7}{\mathrm{m}}^2$; resistivity, $\rho =$ ? Now, or $\rho =\frac{\mathrm{RA}}{\mathrm{L}}=\frac{50\times 6.0\times 10^{-7}}{15}=2\times {10}^{-6}\Omega \mathbf{m}$
3. A wire of resistance $5\Omega$ is drawn out so that its length is increased by twice its original length. Calculate its new resistance. Solution: Let the initial resistance, initial length and initial area of cross-section of the wire are ${\mathrm{R}}_1,{\mathrm{L}}_1$ and ${\mathrm{A}}_1$ respectively. Let the final resistance, final length and final area of cross-section of the wire are ${\mathrm{R}}_2,{\mathrm{L}}_2$ and ${\mathrm{A}}_2$ respectively. Given, ${\mathrm{R}}_1=5\Omega$ Now, the wire is drawn out so that its length is increased by twice its original length, that is, ${\mathrm{L}}_2={\mathrm{L}}_1+2{\mathrm{L}}_1=3{\mathrm{L}}_1$ Now, initial volume $=$ final volume That is, $A_1{L}_1=A_2{L}_2$ or $A_1{L}_1=A_2\left(3L_1\right)$ or $A_2=\frac{A_1}{3}$ Now, initial resistance, ${\mathrm{R}}_1=\rho \frac{{\mathrm{L}}_1}{{\mathrm{A}}_1}$ Final resistance, $R_2=\rho \frac{L_2}{A_2}=\rho \frac{3L_1}{\left(A_1/3\right)}=(3{)}^2\left(\rho \frac{L_1}{A_1}\right)$ or ${\mathrm{R}}_2=9{\mathrm{R}}_1=9\times 5=45\Omega$
4. Consider the circuit shown in the diagram. (a) If $\epsilon =12\mathrm{V},{\mathrm{R}}_1=10\mathrm{k}\Omega ,{\mathrm{R}}_2=5\mathrm{k}\Omega$ and ${\mathrm{R}}_3=8\mathrm{k}\Omega$, calculate the voltage across ${\mathrm{R}}_2$. (b) If $\epsilon =10\mathrm{V},{\mathrm{V}}_1=3\mathrm{V},{\mathrm{R}}_2=10\mathrm{k}\Omega$ and ${\mathrm{R}}_3=4\mathrm{k}\Omega$, calculate the resistance of R1.
   
   Solution: (a) The total resistance of the series circuit, ${\mathrm{R}}_{\mathrm{e}}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3=10+5+8=23\mathrm{k}\Omega$ The current flowing around the circuit, $\mathrm{I}=\frac{\epsilon }{\mathrm{R}}=\frac{12}{23\times 10^3}=5.2\times 10^{-4}\mathrm{A}$ Hence the voltage across ${\mathrm{R}}_2$, ${\mathrm{V}}_2=\mathrm{I}\times {\mathrm{R}}_2$ $=\left(5.2\times 10^{-4}\right)\times \left(5\times 10^3\right)=2.6\mathbf{V}$ (b) The total voltage across the resistors ${\mathrm{R}}_2$ and ${\mathrm{R}}_3,\mathrm{V}=\epsilon -{\mathrm{V}}_1=10-3=7\mathrm{V}$ Now, $R_2$ and $R_3$ are in series, their total resistance is $\mathrm{R}={\mathrm{R}}_2+{\mathrm{R}}_3=10+4=14\mathrm{k}\Omega$ Hence the current flowing around the circuit, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{7}{14\times 10^3}=5\times 10^{-4}\mathrm{A}$ Now, the resistance of $R_1$ is $$\begin{gathered} {\mathrm{R}}_1=\frac{{\mathrm{V}}_1}{\mathrm{I}}=\frac{3}{5\times 10^{-4}}=6\times 10^3\Omega \\ =6k\mathbf{\Omega } \end{gathered}$$
5. The resistances of two conductors in series is $40\Omega$ and their resistance becomes $6.4\Omega$ when connected in parallel. Find the individual resistances. Solution: Let $R_1,R_2$ be the two resistances of conductors. When they are connected in series, their total resistance, $R_1+R_2=40\Omega$ When they are connected in parallel, their total resistance is $\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}=6.4$ or $\frac{R_1{R}_2}{40}=6.4$ [Using (1)] or ${\mathrm{R}}_1{\mathrm{R}}_2=6.4\times 40=256$ or $R_1=256/R_2$ Putting the value of $R_1$ from (2) in (1), we get $\frac{256}{{\mathrm{R}}_2}+{\mathrm{R}}_2=40\text{ or }\frac{256+{\mathrm{R}}_2^2}{{\mathrm{R}}_2}=40$ or $R_2^2-40R_2+256=0$ or $R_2^2-32R_2-8R_2+256=0$ or $R_2\left(R_2-32\right)-8\left(R_2-32\right)=0$ or $\left(R_2-8\right)\left(R_2-32\right)=0$ or ${\mathrm{R}}_2=8\Omega$ and ${\mathrm{R}}_2=32\Omega$ When ${\mathrm{R}}_2=8\Omega ,{\mathrm{R}}_1=40-8=32\Omega$ When ${\mathrm{R}}_2=32\Omega ,{\mathrm{R}}_1=40-32=8\Omega$ Thus, the two resistors have resistances $8\Omega$ and $32\Omega$ respectively.
6. Find the value of the current in the circuit shown below.
   
   Solution: Resistors AB and BC are in series, their equivalent resistance, ${\mathrm{R}}_1=30+30=60\Omega$ Now, $R_1$ and the resistor $AC$ are in parallel, their equivalent resistance, $$\begin{gathered} R_e=\frac{R_1\times 30}{R_1+30}=\frac{60\times 30}{60+30}=\frac{60\times 30}{90} \\ =20\Omega \end{gathered}$$ Current in the circuit, $\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{e}}}=\frac{2}{20}=0.1\mathrm{A}$
7. Given, the three resistors of resistances $1\Omega ,2\Omega ,3\Omega$. Make any five possible combinations from these resistors. Find equivalent resistance in each case. Solution: (i) $1\Omega ,2\Omega ,3\Omega$ all joined in series. Their equivalent resistance, ${\mathrm{R}}_{\mathrm{e}}=1+2+3=6\mathbf{\Omega }$
   
   (ii) $1\Omega ,2\Omega ,3\Omega$ all joined in parallel. Their equivalent resistance, $\frac{1}{{\mathrm{R}}_{\mathrm{e}}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6}$
   
   or $R_e=\frac{11}{6}\Omega$ (iii) $1\Omega ,2\Omega$ are joined in parallel, their combination is joined in series with $3\Omega$. Equivalent resistance of $1\Omega ,2\Omega$, ${\mathrm{R}}_1=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega$
   
   Now, ${\mathrm{R}}_1$ and $3\Omega$ are joined in series, their equivalent resistance, ${\mathrm{R}}_{\mathrm{e}}={\mathrm{R}}_1+3=\frac{2}{3}+3=\frac{2+9}{3}=\frac{11}{3}\mathbf{\Omega }$ (iv) $2\Omega ,3\Omega$ are joined in parallel, their combination is joined in series with $1\Omega$. Equivalent resistance of $2\Omega ,3\Omega$, ${\mathrm{R}}_1=\frac{2\times 3}{2+3}=\frac{6}{5}\Omega$
   
   Now, ${\mathrm{R}}_1$ and $1\Omega$ are joined in series, their equivalent resistance, ${\mathrm{R}}_{\mathrm{e}}={\mathrm{R}}_1+1=\frac{6}{5}+1=\frac{6+5}{5}=\frac{11}{5}\mathbf{\Omega }$ (v) $1\Omega ,3\Omega$ are joined in parallel, their combination is joined in series with $2\Omega$. Equivalent resistance of $1\Omega ,3\Omega$, ${\mathrm{R}}_1=\frac{1\times 3}{1+3}=\frac{3}{4}\Omega$
   
   Now, ${\mathrm{R}}_1$ and $2\Omega$ are joined in series, their equivalent resistance, ${\mathrm{R}}_{\mathrm{e}}={\mathrm{R}}_1+2=\frac{3}{4}+2=\frac{3+8}{4}=\frac{11}{4}\mathbf{\Omega }$
8. Consider the circuit shown in the diagram. (a) If $\epsilon =15\mathrm{V},{\mathrm{R}}_1=10\mathrm{k}\Omega ,{\mathrm{R}}_2=5\mathrm{k}\Omega$ and ${\mathrm{R}}_3=8\mathrm{k}\Omega$, calculate the total current flowing through the circuit. (b) If $\epsilon =10\mathrm{V},\mathrm{I}=10\mathrm{mA},{\mathrm{R}}_2=10\mathrm{k}\Omega$ and ${\mathrm{R}}_3=4\mathrm{k}\Omega$, calculate the resistance of R1.
   
   Solution (a) The total resistance of the parallel circuit, $\frac{1}{{\mathrm{R}}_{\mathrm{e}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}=\frac{1}{10}+\frac{1}{5}+\frac{1}{8}=\frac{4+8+5}{40}$ or $\frac{1}{{\mathrm{R}}_{\mathrm{e}}}=\frac{17}{40}$ or ${\mathrm{R}}_{\mathrm{e}}=\frac{40}{17}\mathrm{k}\Omega$ or $R_e=(40/17)\times 10^3\Omega$ Hence the total current flowing around the circuit, $\mathrm{I}=\frac{\epsilon }{{\mathrm{R}}_{\mathrm{e}}}=\frac{15}{(40/17)\times 10^3}=\frac{15\times 17}{40\times 10^3}$ $=6.375\times 10^{-3}\mathrm{A}$ (b) The current through ${\mathrm{R}}_2$, ${\mathrm{I}}_2=\frac{\epsilon }{{\mathrm{R}}_2}=\frac{10}{10\times 10^3}=1\times 10^{-3}\mathrm{A}=1\mathrm{mA}$ Similarly, current through ${\mathrm{R}}_3$, $$\begin{gathered} {\mathrm{I}}_3=\frac{\epsilon }{{\mathrm{R}}_3}=\frac{10}{4\times 10^3} \\ =2.5\times 10^{-3}\mathrm{A} \\ =2.5\mathrm{mA} \end{gathered}$$ Thus, the current flowing through ${\mathrm{R}}_1$, $$\begin{gathered} {\mathrm{I}}_1=\mathrm{I}-\left({\mathrm{I}}_2+{\mathrm{I}}_3\right) \\ =10-(1+2.5) \\ =10-3.5=6.5\mathrm{mA} \end{gathered}$$ Hence the resistance of $R_1$ is ${\mathrm{R}}_1=\frac{\epsilon }{{\mathrm{I}}_1}=\frac{10}{6.5\times 10^{-3}}=1538\Omega$ $=1.538\times 10^3\Omega$