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      Class 12 PHYSICS – JEE

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      • Class 12
      • Class 12 PHYSICS – JEE
      CoursesClass 12PhysicsClass 12 PHYSICS – JEE
      • 1.Electrostatics (1)
        8
        • Lecture1.1
          Charge, Coulombs Law and Coulombs law in Vector form 41 min
        • Lecture1.2
          Electric Field; Electric Field Lines; Field lines due to multiple charges 42 min
        • Lecture1.3
          Charge Distribution; Finding Electric Field due to Different Object 01 hour
        • Lecture1.4
          Solid angle; Area Vector; Electric Flux; Flux of closed surface; Gauss Law 47 min
        • Lecture1.5
          Finding E Using Concept of Gauss law and Flux 01 hour
        • Lecture1.6
          Chapter Notes – Electrostatics (1)
        • Lecture1.7
          NCERT Solutions – Electrostatics
        • Lecture1.8
          Revision Notes Electrostatics
      • 2.Electrostatics (2)
        7
        • Lecture2.1
          Work done by Electrostatic Force; Work done by man in E-Field; Electrostatic Potential Energy 49 min
        • Lecture2.2
          Finding Electric Potential, Equipotential Surface and Motion in Electric Field 01 hour
        • Lecture2.3
          Electric Dipole and Dipole in Uniform and Non-uniform Electric field 01 hour
        • Lecture2.4
          Analysis of charge on conductors; Potential due to induced charge 58 min
        • Lecture2.5
          Conductors with cavity- Case 1: Empty cavity, Case 2: Charge Inside Cavity 41 min
        • Lecture2.6
          Connecting Two Conductors; Grounding of conductor; Electric field just outside conductor; Electrostatic pressure; Self potential Energy 54 min
        • Lecture2.7
          Chapter Notes – Electrostatics (2)
      • 3.Current Electricity (1)
        9
        • Lecture3.1
          Current, Motion of Electrons in Conductor; Temp. Dependence of Resistor 26 min
        • Lecture3.2
          Circuit Theory and Kirchoffs Laws 31 min
        • Lecture3.3
          Some Special Circuits- Series & Parallel Circuits, Open Circuit, Short Circuit 26 min
        • Lecture3.4
          Wheatstone Bridge, Current Antisymmetric 21 min
        • Lecture3.5
          Equivalent Resistance- Series and parallel, Equipotential Points, Wheatstone Bridge 25 min
        • Lecture3.6
          Current Antisymmetric, Infinite Ladder, Circuit Solving, 3D circuits 20 min
        • Lecture3.7
          Chapter Notes – Current Electricity
        • Lecture3.8
          NCERT Solutions – Current Electricity
        • Lecture3.9
          Revision Notes Current Electricity
      • 4.Current Electricity (2)
        4
        • Lecture4.1
          Heating Effect of Current; Rating of Bulb; Fuse 19 min
        • Lecture4.2
          Battery, Maximum power theorem; Ohmic and Non Ohmic Resistance; Superconductor 31 min
        • Lecture4.3
          Galvanometer; Ammeter & Voltmeter and Their Making 44 min
        • Lecture4.4
          Potentiometer and its applications ; Meter Bridge; Post Office Box; Colour Code of Resistors 32 min
      • 5.Capacitor
        6
        • Lecture5.1
          Capacitor and Capacitance; Energy in Capacitor 38 min
        • Lecture5.2
          Capacitive Circuits- Kirchoff’s Laws; Heat Production 01 hour
        • Lecture5.3
          Equivalent Capacitance; Charge on both sides of cap. Plate 52 min
        • Lecture5.4
          Dielectric Strength; Polar and Non-Polar Dielectric; Equivalent Cap. with Dielectric 01 hour
        • Lecture5.5
          Inserting and Removing Dielectric- Work (Fringing Effect), Force; Force between plates of capacitor 38 min
        • Lecture5.6
          Revision Notes Capacitor
      • 6.RC Circuits
        3
        • Lecture6.1
          Maths Needed for RC Circuits, RC circuits-Charging Circuit 19 min
        • Lecture6.2
          RC circuits-Discharging Circuit, Initial & Steady State, Final (Steady) State, Internal Resistance of Capacitor 44 min
        • Lecture6.3
          Revision Notes RC Circuits
      • 7.Magnetism and Moving Charge
        16
        • Lecture7.1
          Introduction, Vector Product, Force Applied by Magnetic Field, Lorentz Force, Velocity Selector 40 min
        • Lecture7.2
          Motion of Charged Particles in Uniform Magnetic Field 40 min
        • Lecture7.3
          Cases of Motion of Charged Particles in Uniform Magnetic Field 56 min
        • Lecture7.4
          Force on a Current Carrying Wire on Uniform B and its Cases, Questions and Solutions 59 min
        • Lecture7.5
          Magnetic Field on Axis of Circular Loop, Magnetic field due to Moving Charge, Magnetic Field due to Current 52 min
        • Lecture7.6
          Magnetic Field due to Straight Wire, Different Methods 40 min
        • Lecture7.7
          Magnetic Field due to Rotating Ring and Spiral 41 min
        • Lecture7.8
          Force between Two Current Carrying Wires 36 min
        • Lecture7.9
          Force between Two Current Carrying Wires 58 min
        • Lecture7.10
          Miscellaneous Questions 55 min
        • Lecture7.11
          Solenoid, Toroid, Magnetic Dipole, Magnetic Dipole Momentum, Magnetic Field of Dipole 54 min
        • Lecture7.12
          Magnetic Dipole in Uniform Magnetic Field, Moving Coil Galvanometer, Torsional Pendulum 01 hour
        • Lecture7.13
          Advanced Questions, Magnetic Dipole and Angular Momentum 56 min
        • Lecture7.14
          Chapter Notes – Magnetism and Moving Charge
        • Lecture7.15
          NCERT Solutions – Magnetism and Moving Charge
        • Lecture7.16
          Revision Notes Magnetism and Moving Charge
      • 8.Magnetism and Matter
        10
        • Lecture8.1
          Magnetic Dipole, Magnetic Properties of Matter, Diamagnetism; Domain Theory of Ferro 47 min
        • Lecture8.2
          Magnetic Properties of Matter in Detail 39 min
        • Lecture8.3
          Magnetization and Magnetic Intensity, Meissner Effect, Variation of Magnetization with Temperature 55 min
        • Lecture8.4
          Hysteresis, Permanent Magnet, Properties of Ferro for Permanent Magnet, Electromagnet 31 min
        • Lecture8.5
          Magnetic Compass, Earth’s Magnetic Field 20 min
        • Lecture8.6
          Bar Magnet, Bar Magnet in Uniform Field 49 min
        • Lecture8.7
          Magnetic Poles, Magnetic Field Lines, Magnetism and Gauss’s Law 32 min
        • Lecture8.8
          Chapter Notes – Magnetism and Matter
        • Lecture8.9
          NCERT Solutions – Magnetism and Matter
        • Lecture8.10
          Revision Notes Magnetism and Matter
      • 9.Electromagnetic Induction
        14
        • Lecture9.1
          Introduction, Magnetic Flux, Motional EMF 01 min
        • Lecture9.2
          Induced Electric Field, Faraday’s Law, Comparison between Electrostatic Electric Field and Induced Electric Field 43 min
        • Lecture9.3
          Induced Current; Faraday’s Law ; Lenz’s Law 56 min
        • Lecture9.4
          Faraday’s Law and its Cases 50 min
        • Lecture9.5
          Advanced Questions on Faraday’s Law 37 min
        • Lecture9.6
          Cases of Current Electricity 59 min
        • Lecture9.7
          Lenz’s Law and Conservation of Energy, Eddy Current, AC Generator, Motor 01 hour
        • Lecture9.8
          Mutual Induction 53 min
        • Lecture9.9
          Self Inductance, Energy in an Inductor 34 min
        • Lecture9.10
          LR Circuit, Decay Circuit 01 hour
        • Lecture9.11
          Initial and Final Analysis of LR Circuit 38 min
        • Lecture9.12
          Chapter Notes – Electromagnetic Induction
        • Lecture9.13
          NCERT Solutions – Electromagnetic Induction
        • Lecture9.14
          Revision Notes Electromagnetic Induction
      • 10.Alternating Current Circuit
        8
        • Lecture10.1
          Introduction, AC/DC Sources, Basic AC Circuits, Average & RMS Value 46 min
        • Lecture10.2
          Phasor Method, Rotating Vector, Adding Phasors, RC Circuit 35 min
        • Lecture10.3
          Examples and Solutions 21 min
        • Lecture10.4
          Power in AC Circuit, Resonance Frequency, Bandwidth and Quality Factor, Transformer 51 min
        • Lecture10.5
          LC Oscillator, Question and Solutions of LC Oscillator, Damped LC Oscillator 53 min
        • Lecture10.6
          Chapter Notes – Alternating Current Circuit
        • Lecture10.7
          NCERT Solutions – Alternating Current Circuit
        • Lecture10.8
          Revision Notes Alternating Current Circuit
      • 11.Electromagnetic Waves
        4
        • Lecture11.1
          Displacement Current; Ampere Maxwell Law 14 min
        • Lecture11.2
          EM Waves; EM Spectrum; Green House Effect; Ozone Layer 36 min
        • Lecture11.3
          Chapter Notes – Electromagnetic Waves
        • Lecture11.4
          Revision Notes Electromagnetic Waves
      • 12.Photoelectric Effect
        5
        • Lecture12.1
          Recalling Basics; Photoelectric Effect 50 min
        • Lecture12.2
          Photo-electric Cell 35 min
        • Lecture12.3
          Photon Flux; Photon Density; Momentum of Photon; Radiation Pressure- Full Absorption, Full Reflection; Dual nature 52 min
        • Lecture12.4
          Chapter Notes – Photoelectric Effect
        • Lecture12.5
          Revision Notes Photoelectric Effect
      • 13.Ray Optics (Part 1)
        12
        • Lecture13.1
          Rays and Beam of Light, Reflection of Light, Angle of Deviation, Image Formation by Plane Mirror 01 hour
        • Lecture13.2
          Field of View, Numerical on Field of Line, Size of Mirror 42 min
        • Lecture13.3
          Curved Mirrors, Terms Related to Curved Mirror, Reflection of Light by Curved Mirror 40 min
        • Lecture13.4
          Image Formation by Concave Mirror, Magnification or Lateral or Transverse Magnification 01 hour
        • Lecture13.5
          Ray Diagrams for Concave Mirror 45 min
        • Lecture13.6
          Image Formation by Convex Mirror; Derivations of Various Formulae used in Concave Mirror and Convex Mirror 01 hour
        • Lecture13.7
          Advanced Optical Systems, Formation of Images with more than one Mirror 24 min
        • Lecture13.8
          Concept of Virtual Object, Formation of Image when Incident ray are Converging, Image Characteristics for Virtual Object, 55 min
        • Lecture13.9
          Newton’s Formula, Longitudinal Magnification 23 min
        • Lecture13.10
          Formation of Image when Two Plane Mirrors kept at an angle and parallel; Formation of Image by two Parallel Mirrors. 43 min
        • Lecture13.11
          Chapter Notes – Ray Optics
        • Lecture13.12
          NCERT Solutions – Ray Optics
      • 14.Ray Optics (Part 2)
        13
        • Lecture14.1
          Refractive Index, Opaque, Transparent, Speed of Light, Relative Refractive Index, Refraction and Snell’s Law, Refraction in Denser and Rarer Medium 42 min
        • Lecture14.2
          Image Formation due to Refraction; Derivation; Refraction and Image formation in Glass Slab 57 min
        • Lecture14.3
          Total Internal Reflection, Critical Angle, Principle of Reversibility 01 hour
        • Lecture14.4
          Application of Total Internal Reflection 45 min
        • Lecture14.5
          Refraction at Curved Surface, Image Formation by Curved Surface, Derivation 56 min
        • Lecture14.6
          Image Formation by Curved Surface, Snell’s Law in Vector Form 01 hour
        • Lecture14.7
          Lens, Various types of Lens, Differentiating between various Lenses; Optical Centre, Derivation of Lens Maker Formula 01 hour
        • Lecture14.8
          Lens Formula, Questions and Answers 39 min
        • Lecture14.9
          Property of Image by Convex and Concave Lens; Lens Location, Minimum Distance Between Real Image and Object 01 hour
        • Lecture14.10
          Power of Lens, Combination of Lens, Autocollimation 35 min
        • Lecture14.11
          Silvering of Lens 44 min
        • Lecture14.12
          Cutting of Lens and Mirror, Vertical Cutting, Horizontal Cutting 49 min
        • Lecture14.13
          Newton’s Law for Lens and Virtual Object 01 hour
      • 15.Ray Optics (Part 3)
        6
        • Lecture15.1
          Prism, Angle of Prism, Reversibility in Prism 51 min
        • Lecture15.2
          Deviation in Prism, Minimum and Maximum Deviation, Asymmetric, Thin Prism, Proof for formula of Thin Prism 59 min
        • Lecture15.3
          Dispersion of Light, Refractive Index, Composition of Light, Dispersion through Prism 01 hour
        • Lecture15.4
          Rainbow Formation, Scattering of Light, Tyndall Effect, Defects of Image, Spherical Defect, Chromatic Defect, Achromatism. 57 min
        • Lecture15.5
          Optical Instruments, The Human Eye, Defects of Eye and its Corrections 01 hour
        • Lecture15.6
          Microscope & Telescope 02 hour
      • 16.Wave Optics
        21
        • Lecture16.1
          Introduction to Wave Optics 11 min
        • Lecture16.2
          Huygens Wave Theory 14 min
        • Lecture16.3
          Huygens Theory of Secondary Wavelets 10 min
        • Lecture16.4
          Law of Reflection by Huygens Theory 10 min
        • Lecture16.5
          Deriving Laws of Refraction by Huygens Wave Theory 10 min
        • Lecture16.6
          Multiple Answer type question on Huygens Theory 41 min
        • Lecture16.7
          Conditions of Constructive and Destructive Interference 22 min
        • Lecture16.8
          Conditions of Constructive and Destructive Interference 06 min
        • Lecture16.9
          Conditions of Constructive and Destructive Interference 23 min
        • Lecture16.10
          Incoherent Sources of Light 38 min
        • Lecture16.11
          Youngs Double Slit Experiment 12 min
        • Lecture16.12
          Fringe Width Positions of Bright and Dark Fringes 15 min
        • Lecture16.13
          Numerical problems on Youngs Double Slit Experiment 11 min
        • Lecture16.14
          Numerical problems on Youngs Double Slit Experiment 19 min
        • Lecture16.15
          Displacement of Interference Pattern 19 min
        • Lecture16.16
          Numerical problems on Displacement of Interference Pattern 28 min
        • Lecture16.17
          Shapes of Fringes 37 min
        • Lecture16.18
          Colour of Thin Films 59 min
        • Lecture16.19
          Interference with White Light 32 min
        • Lecture16.20
          Chapter Notes – Wave Optics
        • Lecture16.21
          NCERT Solutions – Wave Optics
      • 17.Atomic Structure
        6
        • Lecture17.1
          Thomson and Rutherford Model of Atom; Trajectory of Alpha particle; Bohr’s Model ; Hydrogen Like Atom; Energy Levels 58 min
        • Lecture17.2
          Emission Spectra, Absorption Spectra; De Broglie Explanation of Bohr’s 2nd Postulate; Limitations of Bohr’s Model 37 min
        • Lecture17.3
          Momentum Conservation in Photon Emission, Motion of Nucleus, Atomic Collision 58 min
        • Lecture17.4
          Chapter Notes – Atomic Structure
        • Lecture17.5
          NCERT Solutions – Atomic Structure
        • Lecture17.6
          Revision Notes Atomic Structure
      • 18.Nucleus
        6
        • Lecture18.1
          Basics- Size of Nucleus, Nuclear Force, Binding Energy, Mass Defect; Radioactive Decay 01 hour
        • Lecture18.2
          Laws of Radioactive Decay 36 min
        • Lecture18.3
          Nuclear Fission; Nuclear Reactor; Nuclear Fusion- Reaction Inside Sun 30 min
        • Lecture18.4
          Chapter Notes – Nucleus
        • Lecture18.5
          NCERT Solutions – Nucleus
        • Lecture18.6
          Revision Notes Nucleus
      • 19.X-Ray
        4
        • Lecture19.1
          Electromagnetic Spectrum, Thermionic Emission; Coolidge Tube – Process 1 22 min
        • Lecture19.2
          Coolidge Tube – Process 2; Moseley’s Law; Absorption of X-rays in Heavy Metal 39 min
        • Lecture19.3
          Chapter Notes – X-Ray
        • Lecture19.4
          Revision Notes X-Ray
      • 20.Error and Measurement
        2
        • Lecture20.1
          Least Count of Instruments; Mathematical Operation on Data with Random Error 18 min
        • Lecture20.2
          Significant Digits; Significant Digits and Mathematical Operations 30 min
      • 21.Semiconductors
        9
        • Lecture21.1
          Conductor, Semiconductors and Insulators Basics Difference, Energy Band Theory, Si element 21 min
        • Lecture21.2
          Doping and PN Junction 01 hour
        • Lecture21.3
          Diode and Diode as Rectifier 01 hour
        • Lecture21.4
          Voltage Regulator and Zener Diode and Optoelectronic Jn. Devices 01 hour
        • Lecture21.5
          Transistor, pnp, npn, Modes of operation, Input and Output Characteristics, , Current Amplification Factor 01 hour
        • Lecture21.6
          Transistor as Amplifier, Transistor as Switch, Transistor as Oscillator, Digital Gates 01 hour
        • Lecture21.7
          Chapter Notes – Semiconductors
        • Lecture21.8
          NCERT Solutions – Semiconductors
        • Lecture21.9
          Revision Notes Semiconductors
      • 22.Communication Systems
        5
        • Lecture22.1
          Basic working and terms; Antenna; Modulation and Types of Modulation 32 min
        • Lecture22.2
          Amplification Modulation, Transmitter, Receiver, Modulation index 40 min
        • Lecture22.3
          Chapter Notes – Communication Systems
        • Lecture22.4
          NCERT Solutions – Communication Systems
        • Lecture22.5
          Revision Notes Communication Systems

        NCERT Solutions – Current Electricity

        3.1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?

        ANSWER:

        Emf of the battery, E = 12 V

        Internal resistance of the battery, r = 0.4 Ω

        Maximum current drawn from the battery = I

        According to Ohm’s law,

        The maximum current drawn from the given battery is 30 A.

        3.2. A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

        ANSWER:

        Emf of the battery, E = 10 V

        Internal resistance of the battery, r = 3 Ω

        Current in the circuit, I = 0.5 A

        Resistance of the resistor = R

        The relation for current using Ohm’s law is,

        Terminal voltage of the resistor = V

        According to Ohm’s law,

        V = IR

        = 0.5 × 17

        = 8.5 V

        Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

        8.5 V.

         3.3.

        (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

        (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

        ANSWER:

        (a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.

        Total resistance = 1 + 2 + 3 = 6 Ω

        (b) Current flowing through the circuit = I

        Emf of the battery, E = 12 V

        Total resistance of the circuit, R = 6 Ω

        The relation for current using Ohm’s law is,

        Potential drop across 1 Ω resistor = V1

        From Ohm’s law, the value of V1 can be obtained as

        V1 = 2 × 1= 2 V … (i)

        Potential drop across 2 Ω resistor = V2

        Again, from Ohm’s law, the value of V2 can be obtained as

        V2 = 2 × 2= 4 V … (ii)

        Potential drop across 3 Ω resistor = V3

        Again, from Ohm’s law, the value of V3 can be obtained as

        V3 = 2 × 3= 6 V … (iii)

        Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.

        3.4.

        (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

        (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

        ANSWER:

        (a) There are three resistors of resistances,

        R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω

        They are connected in parallel. Hence, total resistance (R) of the combination is given by,

        Therefore, total resistance of the combination is.

        (b) Emf of the battery, V = 20 V

        Current (I1) flowing through resistor R1 is given by,

        Current (I2) flowing through resistor R2 is given by,

        Current (I3) flowing through resistor R3 is given by,

        Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A

        Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.

        3.5. At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 

        ANSWER:

        Room temperature, T = 27°C

        Resistance of the heating element at T, R = 100 Ω

        Let T1 is the increased temperature of the filament.

        Resistance of the heating element at T1, R1 = 117 Ω

        Temperature co-efficient of the material of the filament,

        Therefore, at 1027°C, the resistance of the element is 117Ω.

        3.6.A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

        ANSWER:

        Length of the wire, l =15 m

        Area of cross-section of the wire, a = 6.0 × 10−7 m2

        Resistance of the material of the wire, R = 5.0 Ω

        Resistivity of the material of the wire = ρ

        Resistance is related with the resistivity as

        Therefore, the resistivity of the material is 2 × 10−7 Ω m.

        3.7. A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

        ANSWER:

        Temperature, T1 = 27.5°C

        Resistance of the silver wire at T1, R1 = 2.1 Ω

        Temperature, T2 = 100°C

        Resistance of the silver wire at T2, R2 = 2.7 Ω

        Temperature coefficient of silver = α

        It is related with temperature and resistance as

        Therefore, the temperature coefficient of silver is 0.0039°C−1.

        3.8. Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10−4 °C −1.

        ANSWER:

        Supply voltage, V = 230 V

        Initial current drawn, I1 = 3.2 A

        Initial resistance = R1, which is given by the relation,

        Steady state value of the current, I2 = 2.8 A

        Resistance at the steady state = R2, which is given as

        Temperature co-efficient of nichrome, α = 1.70 × 10−4 °C −1

        Initial temperature of nichrome, T1= 27.0°C

        Study state temperature reached by nichrome = T2

        T2 can be obtained by the relation for α,

        Therefore, the steady temperature of the heating element is 867.5°C

         

        3.9. Determine the current in each branch of the network shown in fig 3.30:

        ANSWER:

        Current flowing through various branches of the circuit is represented in the given figure.

        I1 = Current flowing through the outer circuit

        I2 = Current flowing through branch AB

        I3 = Current flowing through branch AD

        I2 − I4 = Current flowing through branch BC

        I3 + I4 = Current flowing through branch CD

        I4 = Current flowing through branch BD

        For the closed circuit ABDA, potential is zero i.e.,

        10I2 + 5I4 − 5I3 = 0

        2I2 + I4 −I3 = 0

        I3 = 2I2 + I4 … (1)

        For the closed circuit BCDB, potential is zero i.e.,

        5(I2 − I4) − 10(I3 + I4) − 5I4 = 0

        5I2 + 5I4 − 10I3 − 10I4 − 5I4 = 0

        5I2 − 10I3 − 20I4 = 0

        I2 = 2I3 + 4I4 … (2)

        For the closed circuit ABCFEA, potential is zero i.e.,

        −10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

        10 = 15I2 + 10I1 − 5I4

        3I2 + 2I1 − I4 = 2 … (3)

        From equations (1) and (2), we obtain

        I3 = 2(2I3 + 4I4) + I4

        I3 = 4I3 + 8I4 + I4

        − 3I3 = 9I4

        − 3I4 = + I3 … (4)

        Putting equation (4) in equation (1), we obtain

        I3 = 2I2 + I4

        − 4I4 = 2I2

        I2 = − 2I4 … (5)

        It is evident from the given figure that,

        I1 = I3 + I2 … (6)

        Putting equation (6) in equation (1), we obtain

        3I2 +2(I3 + I2) − I4 = 2

        5I2 + 2I3 − I4 = 2 … (7)

        Putting equations (4) and (5) in equation (7), we obtain

        5(−2 I4) + 2(− 3 I4) − I4 = 2

        − 10I4 − 6I4 − I4 = 2

        17I4 = − 2

        Equation (4) reduces to

        I3 = − 3(I4)

        Therefore, current in branch 

        In branch BC = 

        In branch CD = 

        In branch AD 

        In branch BD = 

        Total current = 

         

        3.10. (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

        (b) Determine the balance point of the bridge above if X and Y are interchanged.

        (c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

        ANSWER:

        A metre bridge with resistors X and Y is represented in the given figure.

        (a) Balance point from end A, l1 = 39.5 cm

        Resistance of the resistor Y = 12.5 Ω

        Condition for the balance is given as,

        Therefore, the resistance of resistor X is 8.2 Ω.

        The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

        (b) If X and Y are interchanged, then l1 and 100−l1 get interchanged.

        The balance point of the bridge will be 100−l1 from A.

        100−l1 = 100 − 39.5 = 60.5 cm

        Therefore, the balance point is 60.5 cm from A.

        (c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

        3.11. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

        ANSWER:

        Emf of the storage battery, E = 8.0 V

        Internal resistance of the battery, r = 0.5 Ω

        DC supply voltage, V = 120 V

        Resistance of the resistor, R = 15.5 Ω

        Effective voltage in the circuit = V1

        R is connected to the storage battery in series. Hence, it can be written as

        V1 = V − E

        V1 = 120 − 8 = 112 V

        Current flowing in the circuit = I, which is given by the relation,

        Voltage across resistor R given by the product, IR = 7 × 15.5 = 108.5 V

        DC supply voltage = Terminal voltage of battery + Voltage drop across R

        Terminal voltage of battery = 120 − 108.5 = 11.5 V

        A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

        3.12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

        ANSWER:

        Emf of the cell, E1 = 1.25 V

        Balance point of the potentiometer, l1= 35 cm

        The cell is replaced by another cell of emf E2.

        New balance point of the potentiometer, l2 = 63 cm

        Therefore, emf of the second cell is 2.25V.

        3.13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m−3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10−6 m2 and it is carrying a current of 3.0 A.

        ANSWER:

        Number density of free electrons in a copper conductor, n = 8.5 × 1028 m−3 Length of the copper wire, l = 3.0 m

        Area of cross-section of the wire, A = 2.0 × 10−6 m2

        Current carried by the wire, I = 3.0 A, which is given by the relation,

        I = nAeVd

        Where,

        e = Electric charge = 1.6 × 10−19 C

        Vd = Drift velocity 

        Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 104 s.

        3.14. The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)

        ANSWER:

        Surface charge density of the earth, σ = 10−9 C m−2

        Current over the entire globe, I = 1800 A

        Radius of the earth, r = 6.37 × 106 m

        Surface area of the earth,

        A = 4πr2

        = 4π × (6.37 × 106)2

        = 5.09 × 1014 m2

        Charge on the earth surface,

        q = σ × A

        = 10−9 × 5.09 × 1014

        = 5.09 × 105 C

        Time taken to neutralize the earth’s surface = t

        Current, 

        Therefore, the time taken to neutralize the earth’s surface is 282.77 s.

         

        3.15. (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

        (b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

        ANSWER:

        (a) Number of secondary cells, n = 6

        Emf of each secondary cell, E = 2.0 V

        Internal resistance of each cell, r = 0.015 Ω

        series resistor is connected to the combination of cells.

        Resistance of the resistor, R = 8.5 Ω

        Current drawn from the supply = I, which is given by the relation,

        Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A

        Therefore, the current drawn from the supply is 1.39 A and terminal voltage is

        11.87 A.

        (b) After a long use, emf of the secondary cell, E = 1.9 V

        Internal resistance of the cell, r = 380 Ω

        Hence, maximum current

        Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.

        3.16. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρAl = 2.63 × 10−8 Ω m, ρCu = 1.72 × 10−8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

        ANSWER:

        Resistivity of aluminium, ρAl = 2.63 × 10−8 Ω m

        Relative density of aluminium, d1 = 2.7

        Let l1 be the length of aluminium wire and m1 be its mass.

        Resistance of the aluminium wire = R1

        Area of cross-section of the aluminium wire = A1

        Resistivity of copper, ρCu = 1.72 × 10−8 Ω m

        Relative density of copper, d2 = 8.9

        Let l2 be the length of copper wire and m2 be its mass.

        Resistance of the copper wire = R2

        Area of cross-section of the copper wire = A2

        The two relations can be written as

        It is given that,

        And,

        Mass of the aluminium wire,

        m1 = Volume × Density

        = A1l1 × d1 = A1 l1d1 … (3)

        Mass of the copper wire,

        m2 = Volume × Density

        = A2l2 × d2 = A2 l2d2 … (4)

        Dividing equation (3) by equation (4), we obtain

        It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper.

        Since aluminium is lighter, it is preferred for overhead power cables over copper.

        3.17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?

        Current

        A

        Voltage

        V

        Current

        A

        Voltage

        V

        0.2

        3.94

        3.0

        59.2

        0.4

        7.87

        4.0

        78.8

        0.6

        11.8

        5.0

        98.6

        0.8

        15.7

        6.0

        118.5

        1.0

        19.7

        7.0

        138.2

        2.0

        39.4

        8.0

        158.0

        ANSWER:

        It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.

         3.18. Answer the following questions:

        (a) A steady current flows in a metallic conductor of non-uniform cross- section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

        (b) Is Ohm’s law universally applicable for all conducting elements?

        If not, give examples of elements which do not obey Ohm’s law.

        (c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?

        (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

        ANSWER:

        (a) When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.

        (b) No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.

        (c) According to Ohm’s law, the relation for the potential is V = IR

        Voltage (V) is directly proportional to current (I).

        R is the internal resistance of the source.

        If V is low, then R must be very low, so that high current can be drawn from the source.

        (d) In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.

        3.19. Choose the correct alternative:

        (a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

        (b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

        (c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.

        (d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/103).

        ANSWER:

        (a) Alloys of metals usually have greater resistivity than that of their constituent metals.

        (b) Alloys usually have lower temperature coefficients of resistance than pure metals.

        (c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.

        (d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.

        3.20. (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

        (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

        (c) Determine the equivalent resistance of networks shown in Fig. 3.31.

        ANSWER:

        (a) Total number of resistors = n

        Resistance of each resistor = R

        (i) When n resistors are connected in series, effective resistance R1is the maximum, given by the product nR.

        Hence, maximum resistance of the combination, R1 = nR

        (ii) When n resistors are connected in parallel, the effective resistance (R2) is the minimum, given by the ratio.

        Hence, minimum resistance of the combination, R2 = 

        (iii) The ratio of the maximum to the minimum resistance is,

        (b) The resistance of the given resistors is,

        R1 = 1 Ω, R2 = 2 Ω, R3 = 3 Ω2

        1. Equivalent resistance, 

        Consider the following combination of the resistors.

        Equivalent resistance of the circuit is given by,

        1. Equivalent resistance, 

        Consider the following combination of the resistors.

        Equivalent resistance of the circuit is given by,

        (iii) Equivalent resistance, R’ = 6 Ω

        Consider the series combination of the resistors, as shown in the given circuit.

        Equivalent resistance of the circuit is given by the sum,

        R’ = 1 + 2 + 3 = 6 Ω

        (iv) Equivalent resistance, 

        Consider the series combination of the resistors, as shown in the given circuit.

        Equivalent resistance of the circuit is given by,

        (c) (a) It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in series.

        Hence, their equivalent resistance = (1+1) = 2 Ω

        It can also be observed that two resistors of resistance 2 Ω each are connected in series.

        Hence, their equivalent resistance = (2 + 2) = 4 Ω.

        Therefore, the circuit can be redrawn as

        It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four loops. Hence, equivalent resistance (R’) of each loop is given by,

        The circuit reduces to,

        All the four resistors are connected in series.

        Hence, equivalent resistance of the given circuit is 

        (b) It can be observed from the given circuit that five resistors of resistance R each are connected in series.

        Hence, equivalent resistance of the circuit = R + R + R + R + R

        = 5 R

        2

         

        3.21. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.

        ANSWER:

        The resistance of each resistor connected in the given circuit, R = 1 Ω

        Equivalent resistance of the given circuit = R’

        The network is infinite. Hence, equivalent resistance is given by the relation,

        Negative value of R’ cannot be accepted. Hence, equivalent resistance,

        Internal resistance of the circuit, r = 0.5 Ω

        Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

        Supply voltage, V = 12 V

        According to Ohm’s Law, current drawn from the source is given by the ratio,  = 3.72 A

        3.22. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

        (a) What is the value ε ?

        (b) What purpose does the high resistance of 600 kΩ have?

        (c) Is the balance point affected by this high resistance?

        (d) Is the balance point affected by the internal resistance of the driver cell?

        (e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?

        (f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

        ANSWER:

        (a) Constant emf of the given standard cell, E1 = 1.02 V

        Balance point on the wire, l1 = 67.3 cm

        A cell of unknown emf, ε,replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm

        The relation connecting emf and balance point is,


        The value of unknown emfis 1.247 V.

        (b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

        (c) The balance point is not affected by the presence of high resistance.

        (d) The point is not affected by the internal resistance of the driver cell.

        (e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

        (f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

        The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

        3.23. Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?

        ANSWER:

        Resistance of the standard resistor, R = 10.0 Ω

        Balance point for this resistance, l1 = 58.3 cm

        Current in the potentiometer wire = i

        Hence, potential drop across R, E1 = iR

        Resistance of the unknown resistor = X

        Balance point for this resistor, l2 = 68.5 cm

        Hence, potential drop across X, E2 = iX

        The relation connecting emf and balance point is,

        Therefore, the value of the unknown resistance, X, is 11.75 Ω.

        If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.

        3.24. Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

        ANSWER:

        Internal resistance of the cell = r

        Balance point of the cell in open circuit, l1 = 76.3 cm

        An external resistance (R) is connected to the circuit with R = 9.5 Ω

        New balance point of the circuit, l2 = 64.8 cm

        Current flowing through the circuit = I

        The relation connecting resistance and emf is,

        Therefore, the internal resistance of the cell is 1.68Ω.

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