1. ELECTRIC CURRENT:-                                                                

Electric current is defined as the flow of charge through a given area of the substance.

Now let small charge passes through a given area than the instantaneous current can be given as

Direction of electric current:-

The direction of flow of positive charge or the direction in which electric field is applied gives the direction of electric current. This current is called conventional current.

The direction of flow of electron gives direction of electronic current.

Unit of electric current: – S.I unit of electric current is Ampere (A)

I.e.        

Current through a conductor is said to be 1 ampere if one coulomb charge flows through any cross sectional area of the conductor in one second.

Electric current is a scalar quantity:-

Electric current has both magnitude as well as direction but it is not a vector quantity. Because it does not follow vector algebra but follow simple (scalar) algebra. The arrow represents the direction of current not that current is a vector quantity.

Types of current:-

Steady direct current (D.C):-

An electric current is said to be steady direct current if its magnitude and direction do not change with time

Alternating Current (A.C):-

An electric current is said to be alternating if its magnitude changes with time and polarity reveres periodically (repeats after a time interval)

Q.1 A current of 1A is drawn by a filament of an electric bulb for 20 minutes. Find the amount of electric charge that flows through the circuit.

Ans: The given data is,   I = 1A and   t = 20 minutes = 20×60 = 1200 seconds

Therefore,              So          = 1200 C

Q.2 A current of 10 A flows through a conductor for two minutes
(i) Calculate the amount of charge passed through any area of cross section of the conductor.
(ii) If the charge of an electron is 1.6 × 10^-19 C, then calculate the total number of electrons flowing.
Answer: Given that: I = 10 A, t = 2 min = 2 × 60 s = 120 s
(i) Amount of charge Q passed through any area of cross-section is given by

(ii) Since, where n is the total number of electrons flowing and e is the charge on one electron
∴   

or   

2 DRIFT velocity m.imp

Drift velocity is defined as the average velocity with which free electrons in a conductor get drifted in a direction opposite to the direction of applied electric field.

(a)Expression for Drift Velocity:-

Let  are the velocities of electrons in a conductor. Then average velocity is                                           

         (1)

If   is the applied electric field then force experienced by charge particles is

        (2)

Note: – negative sign shows that and    are opposite to each other.

If m is the mass of electron than     (3)

From equation (2) and (3)         (4)

The small interval of times between two successive collisions between electrons and ions in the conductor is called relaxation time (τ)

Hence after time τ the Final velocity can be given as ()

 So  Average velocity =  

  =          

Using equation 1 we get      =    

Or             =    

And from equation (2)    =    

Here negative sign show that and are in opposite direction.

• Drift velocity of electron in a conductor is of the order of 10m/s. it is very small compared to the thermal speed which is of the order of
  

  . Drift velocity of electrons in a conductor is of the order of m/s.

(b) Relation between Drift Velocity and Electric Current

Let us consider a conductor of length ℓ and area A. If V is the applied potential difference across the end of conductor of conductor, Then electric field set up can be given as      

As total volume of conductor

If is number of electron per unit volume than total electron

If is charge on 1 electron than charge on electrons

So current flowing in the conductor     

Or     

Hence greater is the drift velocity, greater is the current flow.

Q3.Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area carrying a current of 1.8 A. Assume the density of conduction electrons to be 9 .

Ans: m/s

Q.4 The number density of free electrons in a copper conductor 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.

Ans. Number density of electrons, n = 8.5 × 10²⁸ m⁻³.  l = 3.0 m.   A = 2.0 × 10⁻⁶ m² & I = 3.0 A,

As       =

⟹2.7 × 10⁴ s

 (3) Mobility (μ)

It is the ratio of drift velocity of current carriers and the applied electric field (E).

            (1)

As         =        (2)

From (1) and (2)   

=                      

 (4) Current density:-

Current density of a conductor is defined as the amount of current flowing per unit area of conductor perpendicular to the flow of current density () is a vector quantity.

The current () through an element of surface area     of a conductor is given by

Or                                                                                                                              

 Unit: – The unit of electric current density is Am^-2

(a) Relation between Current Density and Drift Velocity:-

As we know that electric current and drift velocity is related as             

 Or                         

 Hence    

Or     

Hence larger the drift velocity larger will be the current density. 

Q.5 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?

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.                 

 (5) OHM’s – LAW:- ^m.imp

Ohm’s law states that the potential drop across the ends of a conductor of uniform cross section is directly proportional to the current (I) flowing through it provided constant physical condition such as temperature pressure etc.

Mathematically, V  ∝   I 

Or V  =  IR               

Or   R   =  

Where R is constant of proportionality and called electric resistance of the conductor.

     The resistance R not only depends on the material of the conductor But also depends on the dimensions of the conductor.

Proof of ohm’s law:  As we know that current through a conductor is   

 Or   ( τ    )

Or         (E =   )   

                  Or                                              

So by reversing the above equation we get

= constant  

(a) Ohmic and non – Ohmic conductors:-

The conductor which obey ohm’s law called Ohmic conductor (i.e. V-I graph is a straight line) and which do not obey ohm’s law called non – Ohmic conductors (i.e. V-I graph is not straight line).

(b)Limitation/Failure/Exception of Ohm’s Law:-

According to ohm’s law the V-I graph should be straight line but there is some limitation of ohm’s law. These are as follow

(1) Potential Difference may Vary non – linearly with current:-

When current through a conductor increases, the conductor become hot (Temperature increases) due to which the V-I graph does not remain straight line hence ohm’s law fails.

Note:-Always current along y axis and V along x axis but here the result depends upon current so current along x axis and voltage along y axis is taken.

(2)Variation of potential difference may depend upon sign of the potential difference (related to semiconductor physics):-

For P-N junction diode when P-N junction diode is forward biased diode the current flows first slowly than rapidly. Similarly for reverse bias diode.

(3)The Current may be decreasing or increasing the potential difference:-

A thyristor shows this type of behavior .Here the curve AB represents the increase in current with decrease in potential.

(4) In GaAs there is more than one value of V for the same current I

I.e The relation between V and I is not unique.

Q6. A student plots V-I graphs for three samples of nichrome wire with resistances R₁, R₂ and R₃. Choose from the following the statements that holds true for this graph. (2020)
(a) R₁ = R₂ = R₃
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ > R₁
(d) R₂ > R₁ > R₃
Answer:(d) : The inverse of the slope of I-V graph gives the resistance of the material. Here the slope of –R₂ is highest. Thus, R₂ > R₁ > R₃

(6) Resistance:- m.imp

The opposition in the flow of current in a conductor is called resistance of the conductor. Mathematically, it is defined as the ratio of applied potential difference to the current flowing through the conductor. i,e,        

R =          

Unit of Resistance: –   

1 ohm (Ω) = = 1 VA^-1

The resistance of a conductor is said to be 1 ohm if one ampere current flows through it when a potential difference of 1 volt is applied across it.

Cause of Resistance:- When we apply potential difference, the electron starts flowing toward positive end and ions starts vibrating, As flow of electron is affected by ions or we can say the flow of electron is resist by vibrating ions .due to which resistance occurs.

Symbol of Resistance:

     α is positive for conductors.

     α is negative for insulators/semiconductors.

(b)Resistors in Parallel:- 

 Resistor are said to be connected in parallel, if potential difference across each of them is equal to the applied potential difference. Let us consider three resistors are connected in parallel i.e.  

Let I is total current passing through resistor. 

Than           

The potential difference across each resistor is same i.e. 

So               

If  is the equivalent resistance parallel combination 

Than      

or                    

Equivalent Resistance:-

Equivalent Resistance of parallel combination is equal to the sum of reciprocal of individual resistance.

Conclusion:-

     Potential difference across the entire resistor is the same.

     The current through any resistor is inversely proportional to its resistance.

Q25.  A uniform wire of   resistance is bent round in the form of a circle. When connected in a circuit between its two diametrically opposite points, its effective resistance will be:

1. 8Ω                     B. 16Ω                       C. 4Ω                                    D. 2Ω

Q26.  Three resistors of 3 Ω each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery.                     
Answer: equivalent resistance of circuit (R_eq) given by   

   
Using Ohm’s law,  V = IR
We get,    3 V = I × 2 Ω
or   

Q27 . Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases.          
Answer: Let resistance of each resistor be R.    

So, 

For parallel combination,

so required ratio 

Q28.  (a) A 6 Ω resistance wire is doubled on itself. Calculate the new resistance of the wire.
(b) Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer.                                               
Answer: (a) Given, R = 6 Ω          

  = 6 Ω  Now when the length is doubled,

 and 
∴      

(b) Given the total resistance of the combination = 3 Ω

Q29. Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2 V each, a combination of three resistors of 10 Ω, 20 Ω and 30 Ω connected in parallel, a plug key and an ammeter, all connected in series. Use this circuit to find the value of the following: 
(a) Current through each resistor 
(b) Total current in the circuit
(c) Total effective resistance of the circuit. (2020)
Answer: (a) Given,  voltage         =   2V   +   2V   +  2V   =  6 V
Current through 10 Ω      I₁₀ =   V/R   =   6/10   =   0.6 A
Current through 20 Ω       I₂₀ =   V/R  =   6/20   =   0.3 A
Current through 30 Ω        I₃₀ = V/R  =   6/30   =  0.2 A
(b) Total current in the circuit,   1=   I₁₀   +   I₂₀   +   I₃₀   =  0.6   +   0.3   +   0.2   =   1.1 A
(c) Total resistance of the circuit,    

Q30.  Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, an ammeter and a plug key, all connected in series. Also connect a voltmeter to record the potential difference across the 15 Ω resistors and calculate 
(i) the electric current passing through the above circuit and 
(ii) potential difference across 5 Ω resistors when the key is closed. 
Answer: V = (2 × 5) V = 10 V
Equivalent resistance,

R_eq = R₁ + R₂ + R₃ = (5 + 10 + 15)Ω = 30 Ω
(i) Current through circuit,  

I = V/R=10/30 A=0.3 A
(ii) Potential across 5 Ω resistor,

V₁ = IR₁ = 0.3 × 5 = 1.5V 

Q31. 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?

Total number of resistors = n

Resistance of each resistor = R

1.                       When n resistors are connected in series, effective resistance R₁ is the maximum, given by the product nR.
2.                    When n resistors are connected in parallel, the effective resistance (R₂) is the minimum, given by the ratio .

The ratio of the maximum to the minimum resistance is,

Q32. (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 are combined in series.

Total resistance = 1 + 2 + 3 = 6 Ω 

(b) Using Ohm’s law is,

Potential drop across 1 Ω resistor = V₁   

as V₁ = 2 × 1= 2 V … (i) 

Potential drop across 2 Ω resistor = V₂   

as V₂ = 2 × 2= 4 V … (ii) 

Potential drop across 3 Ω resistor = V₃ 

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.

Q33. (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) R₁ = 2 Ω, R₂ = 4 Ω, and R₃ = 5 Ω they are connected in parallel. 

Hence, total resistance    

 (b) V = 20 V so ,  

and  ,  and , 

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

Q34.  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?

Q35. Determine the equivalent resistance of networks shown in Fig

 (A)

 (A)    Ω.                                                                    (b) 5 R

(11) Thermistor:- 

Thermistor is made up of semi-conductor material which is highly heat sensitive. Its resistance change with change in temperature.

Properties of Thermistor:-

     (1)The resistance of Thermistor change with temperature.

     (2)The resistance may decrease or increase with temperature.

Uses:-

(1)They are used for measuring temperature.

(2)They are used for making motors, transformers and generators.

(3)They are used for controlling temperature.

 (12) Super Conductor and Super conductivity:-

The conductors which have zero resistance to electricity is called super conductor. The phenomenon by which a conductor becomes super conductor at a particular temperature (Critical temperature) is called super conductivity.

Critical temperature:-

The temperature at which a substance becomes super conductor is called critical temperature 

Range:- 

At 2-5 ^ok almost all elements act as superconductor.

Critical temperature T_C (K) of some superconducting materials 

Application of superconductor 

1. Superconductor is used in making large electromagnets. This is because no heat is generated when a current flows through a superconductor.

2. Superconductor may be used for transmission of electric power without power losses.

3. Super conductors can be used for making high speed computers. 

(13) A Cell And Related TERMS: – ^m.imp

(a)A Cell:- 

A cell is a device which provided the necessary potential difference to an electric circuit to maintain a continuous flow of current through it.

Symbol of Cell: – Here longer lines show positive terminal and negative by smaller line.

(b)Electromotive Force (EMF)

Emf is the potential difference in a cell when the circuit is open means no current is drawn from cell. Or the work done by a cell to bring a unit +ve charge from one terminal to the other terminal of cell is called emf.

Unit of a cell: – 

Si unit of cell is volt (v) or joule /coulomb.

(c)Terminal potential difference: –  

The potential difference in a closed circuit is called terminal potential difference.

(d)Internal resistance of a call:- 

The resistance offered by electrolyte of the cell is called internal resistance of cell. It is denoted by r.

 (e)Relation between E.M.F and Terminal Potential Difference

 Or. Expression for internal resistance of a cell ^m.imp 

 According to ohms law the current through the circuit is 

     (1)

Or     

Where r is the internal resistance of cell and E is the potential difference when circuit is open.

Potential difference can be written as   

 So by (1)   

⇒          (2)

Which show that EMF is always greater than potential Difference of cell

    If circuit is open         

than from (3)  

Hence if there is no internal resistance then emf is equal to potential difference.

As from equation 2       

or       ( )

 ⇒ 

Q.37 a high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

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.

Q.38  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 E = 12 V,   r = 0.4 Ω   

According to Ohm’s law,   

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

Q. 39 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 E = 10 V  and   r = 3 Ω,  I = 0.5 A 

Using Ohm’s law is

, 

According to Ohm’s law, 

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V.

Q40. A cell of emf E and internal resistance r is connected across a variable resistor R. Plot the shape of graphs showing variation of terminal voltage V with (i) R and (ii) circuit current I.

14. Difference between Emf and Potential difference:-

15 GROUPING OF CELLS

There are three types of grouping of cells.    1 In series   2 in parallel   3 Mixed shells

1.  Cell in series:- add their output voltages, producing a greater voltage

In series combination, the positive of a cell is connected to the positive terminal of the other cell.

Let n cells having each emf E, internal resistance r connected in series to an external resistance R.Then the total emf of cell is  

As the cells are in series so internal resistance are in series 

Here internal resistance and external resistance R is in series so total equivalent resistance  

The current flowing in the circuit is   

 I = 

Special cases:-

1 if  

Than        

So I  

2 If   

than     

So I   

2.  Cell in Parallel:- Batteries last longer in parallel,

Let us consider m identical cells each of emf E and internal resistance r connected in parallel to external resistance R. If cells are connected in parallel 

Than   

Resistance of cells is in parallel so

So      

(iii)Mixed grouping of cells: – ^m.imp

Let us consider a number of identical cells each of emf E and internal resistance r. let these cells be arranged in m rows and each rows contains n cells. This arrangement of cells is connected to an external resistance R.  

Total emf of n cells in row is 

Cells in parallel so 

The internal resistance in each row in series so = nr there are m rows in parallel having each internal resistance, so        

Than total resistance  

The current flowing through the circuit is given by 

  I                                                                

 ⇒   I

The current will be maximum if  are minimum.

i.e.       

Or     R   =  

Thus to obtain maximum current in the circuit, the grouping of cells must be done in such a way that external resistance is equal to effective interval resistance of cells.

Q.41 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 E = 8.0 V, r = 0.5 Ω   V = 120 V, 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 V ¹ = V − E V ¹ = 120 − 8 = 112 V 

Current flowing in the circuit

 = I,   

As  V=

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

Terminal voltage of battery  

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.

16. Kirchhoff’s Laws 

1. Kirchhoff’s I ^st Law (Junction law or current law) ^m.imp

According to Kirchhoff’s I ^st Law in any electric circuit, the algebraic sum of all the current meeting at any junction is zero. Or the sum of current meeting at the junction is equal to the sum of current leaving the junction. 

I, e-     ∑ I=0 

Sign Convention:- 

The current flowing toward Junction is taken  & current leaving the junction are taken. 

In fig. Current  are taken  & are taken  

 As                       

 2. Kirchhoff’s 2 nd Law (loop law or Voltage Law) ^m.imp

According to Kirchhoff’s 2^nd Law, the algebraic sum of the emf in any closed loop is equal to the sum of the product of the current & resistance in it. 

I.e.     ∑ E  = ∑ IR

Sign Convention: –

(1)The emf of a cell is taken, if the direction of traversal is from to terminal.

(2)The emf of a cell is taken, if the direction of traversal is fromto  terminal.

In fig, Apply Kirchhoff’s 2nd Law in loop ABEFA then

Again in loop ABCDEFA  

Or   

Or       ∑E=∑IR

Q42. State Kirchhoff’s rules. Use these rules to write the expressions for the currents   in the circuit diagram shown. 

Ans:  Expressions for the currents   using given loop.

1.   

2. 

3. 

Q43.In the given circuit, with steady current, calculate the potential drop across the capacitor in terms of V.

For loop EBCDE In steady state branch BE is eliminated 

 I

∴

Q44. State Kirchhoff’s rules of current distribution in an electrical network. Using these rules determine the value of the current I1 in the electric circuit given below.

Ans:  Loop ABCFA   ………..(1) 

Loop FCDEF      …….. (2)

At F,      …………. ……..(3)

Solving (1), (2) and (3)   = −0.8A

17. WHEAT STONE BRIDGE ^m.imp

It is an arrangement of four resistances, in which one unknown resistance is determined quickly and accurately in term of three other known resistances. 

Principle:-  

Let S be the unknown resistance, the resistance R is adjusted in such a way that there is no deflection in galvanometer in such situation 

Construction:- 

A wheat stone bridge is consisting of four resistances P, Q, R, and S as four arms of a quadrilateral. A battery of emf E is connected between A& C & a galvanometer between B & D.  As shown in fig.

Proof:-

Applying Kirchhoff’s 2 ^nd Law in loop ABCD

We get   

Again applying loop law in loop BCDB

In balanced condition 

So,    -(i)

&     -(ii) 

Dividing (i) by (ii)   

Hence when there is no deflection in galvanometer then bridge is balanced & unknown resistance can be calculated.              

Sensitivity of a wheat stone bridge :- 

A wheat stone bridge is said to be sensitive if it shows a large deflection in galvanometer for a small change of resistance in the wire. Greater is the sensitivity; larger is the accuracy in the measurement.

Merits/ advantage of Wheatstone Bride:-

(1) The resistance measured by Wheatstone bride does not depend upon internal resistance of battery.

(2)There is no need to measure voltage and current so it is more accurate method because ammeters and voltmeter are not ideal.

(3)The accuracy of resistance measurement can be increased by using high ratio of P and Q. 

18. Metre Bridge or Slide wire bridge: – ^m.imp

It is the simplest measuring unit based on the practical application unit of wheat stone bridge which is used for measuring unknown resistance.

Principle:- 

Its working is based on the principle of wheat stone bridge, i.e. when there is no deflection in galvanometer then bridge is balanced & unknown resistance can be calculated. 

When the bridge is balanced then           

Construction:-  

A metre bridge is consisting of a one metre magnanin or constantan wire fixed on a wooden board along with metre scale with the help of copper strips A& C. Two resistances R (Resistance box) & S are connected between these copper stripes with the help of another copper strips. A Galvanometer is connected with jockey.

Working:- 

When jockey is moved at the wire, then at a certain point B the bridge becomes balanced in that situation,

Where P & Q can be calculated as

Or            

 (here r is the resistance per unit length of the wire)

Or    

So from (1)    

Hence if we know the value of R, l and 100-l than S can be calculated.

Q.45 (a) In a metre bridge 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. 

   Balance point from end A, l1 = 39.5 cm

Resistance of the resistor     Y = 12.5 Ω

Condition for the balance is given as,   

  ⇒X=

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.

19 Heating Effect of Current:

• The effect of electric current due to which heat is produced in a conductor, when current passes through it, is called the heating effect of electric current.

Joule’s law of heating effect:                                              

It states that the heat produced in a resistor is directly proportional to the:

Square of the current in the resistor,   H  ∝  I²       (1)

Resistance of the resistor    H  ∝  R     (2)

Time for which current flows    H  ∝  t       (3)

So,    H    =   I²  R t

• Heating effect is desirable in devices like electric heater, electric iron, electric bulb, electric fuse, etc.
• Heating effect is undesirable in devices like computers, computer monitors (CRT), TV, refrigerators etc.
• In electric bulb, most of the power consumed by the filament appears a heat and a small part of it is radiated in form of light.

Filament of electric bulb is made up of tungsten because:

• It does not oxidize readily at high temperature.
• It has high melting point (3380º C).

The bulbs are filled with chemically inactive gases like nitrogen and argon to prolong the life of filament.

Practical Applications of the Heating Effects of Electric Current

Electrical appliances like laundry iron, toaster, oven, kettle and heater are some devices based on Joule’s Law of Heating. The concept of electric heating is also used to produce light, as in an electric bulb. Another application of Joule’s Law of Heating is the fuse used in electric circuits.

20. Electric Power

• Electric power is the rate at which electrical energy is produced or consumed in an electric circuit

   P = VI = I²R = 

• The S.I. unit of power is watt (W).
• One watt of power is consumed when 1 A of current flows at a potential difference of 1 V. 
• The commercial unit of electric energy is kilowatt hour (kWh), 

Commonly known as a unit 1 kWh = 3.6MJ

Units of Electrical Energy:

The basic unit of electrical energy is the joule or watt-second. An electrical energy is said to be one joule when one ampere of current flows through the circuit for a second when the potential difference of one volt is applied across it. 

The commercial unit of electrical energy is the kilowatt-hour (kWh) which is also known as the Board of trade unit (B.O.T).
 

Generally, one kwh is called one unit.

Q 46. (a) how much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?  

(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?

Solution (a) given V = 220 V; R = 1200 Ω.    

So

 (b) Given, V = 220 V, R = 100 Ω.      

We have 

      Note the difference of current drawn by an electric bulb and electric Heater from the same 220 V source.

Q47 . Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be (2020)
Answer:  Given, P₁ = 100 W and P₂ = 40 W    I₁ = 1 A      I₂ = ?
Since both the bulbs are connected in series, the electric current passing through both the bulbs are same

i.e., I₂ = 1 A.

Q48. Power of a lamp is 60 W. Find the energy in joules consumed by it. 
Answer: Here, P = 60 W &   t = 1 s        

So,    E = Power × time = (60 × 1) J = 60 J

Q49.  Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find current drawn by two bulbs from the line, if the supply voltage is 220 V. 
Answer: Since both the bulbs are connected in parallel and to a 220 V supply,

the voltage across each bulb is 220 V.

And   

Total current drawn from the supply line,   

 I = I₁ + I₂ = 0.454 A + 0.273 A = 0.727 A = 0.73 A

Q50.  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. (2016)
Answer:   Here, V = 220 V, R = 55 Ω
By Ohm’s law    V = IR
   ∴   

Wattage of electric iron = Power 

Q51. An electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb. 

Answer:   Here, 
Power of the bulb 

Q52 . An electric iron has a rating of 750 W; 200 V.

Calculate:(i) the current required.

(ii) the resistance of its heating element.

(iii) Energy consumed by the iron in 2 hours.                               

Answer: Here, P = 750 W, V = 200 V
(i) As    so  

(ii) By Ohm’s law  

or      ∴ 
(iii) Energy consumed by the iron in 2 hours
E

or   

Q53. A bulb is rated 40 W; 220 V. Find the current drawn by it, when it is connected to a 220 V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25 W; 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change. ( 2019)
Answer: In first case, P = 40 W, V = 220 V    

Current drawn 

Also, resistance of bulb,

In second case, P = 25 W, V = 220 V

Current drawn, 

Also, resistance of the bulb,  

Hence, by replacing 40 W bulb to 25 W bulb, having same source of voltage the amount of current flows decreases while resistance increases.

Q54. (a) An electric bulb is connected to a 220 V generator. If the current drawn by the bulb is 0.50 A, find its power 
(b) An electric refrigerator rated 400 W operates 8 hours a day. Calculate the energy per day in kWh.
(c) State the difference between kilowatt and kilowatt hour. (2013)
Answer:  (a) Here, V = 220 V, I = 0.50 A
Power of the bulb, 

(b) Energy consumed by electric refrigerator in a day

(c) Kilowatt is unit of power and kilowatt hour is a unit of energy.

Q55 .(i) State one difference between kilowatt and kilowatt hour. Express 1 kWh in joules.
(ii) A bulb is rated 5V; 500 mA.

Calculate the rated power and resistance of the bulb when it glows. (2013)
Answer:                       

(ii) Here, V = 5 V, I = 500 mA = 0.5 A
Power rating of bulb is    

Resistance of the bulb is 

Some Important Mcq

1. The resistivity of a conductor with increase in temperature increase.
2. With increase in temperature the mobility of electron decreases.
3. With the increase in length of conductor, its resistivity does not change.

(4). If we halve the area of cross section of a conductor resistance of the conductor will be

        (a)  Double         (b)   Half         (c) Four times       (d) One fourth

(5). Electron mobility & resistivity of metallic conductor are related by the relation

         (a)               (b)   ρ ∝μ        (c)              (d)    

(6). If we increase the length of metallic conductor by 2 times resistivity of metallic conductor becomes

          (A)  2 times       (b)   Half          (c)   Remains constant   (d) None of these

(7). S.I. unit of conductivity is

       (a)Ohm               (b)     mho        (c)    Ohm-m                 (d)   

(8). Resistivity of the material is independent of

       (a)Nature of material (b) Temperature of material (c) Dimensions of material   (d) none of these

(9). When electron moves from a wire having a larger area of cross section to a smaller one, the drift velocity of electron

        (a)  Decrease      (b)      Increase    (c)   Remain constant    (d)    None of the above

(10). For insulator, with increase in temperature resistivity wills 25

        (a)    Increase        (b)   Decrease        (c)      Becomes zero        (d)      remains constant

(11). Resistivity of semiconductor with increase in temperature will

        (a)Increase          (b)   Decrease      (c) Remains constant     (d) none of the above

(12). Which of the following is not a unit of electric power

       (a)Joule/sec          (b) Kilowatt hour   (c) Ampere-volt          (d) Watt

(13). A bulb of 4.84 Ω is producing light when connected to 220 v supply what is the electric power of bulb

       (A)  75 watt           (b) 100 watt            (c) 125 watt                (d) 200 watt

Q14. Kirchhoff’s junction rule is based on –

     (a)   Law of conservation of energy         (b) law of conservation of momentum

      (c)  Law of conservation of charge        (d) none of these

Q15. Kirchhoff’s second law is based on law of conservation of –

     (a) Sum of mass & energy     (b) momentum     (c) energy       (d) charge

Q16. For the Kirchhoff’s junction rule which of the following is correct –

     (a) There is no accumulation of charge at junction point (b) algebraic sum of current at junction is zero

     (c) It applies to both open and closed circuit              (d) all of above

Q17. A galvanometer acting as a voltmeter will have with its coil

     (a) A high resistance in parallel       (b) a high resistance in series

     (c) A low resistance in parallel         (d) a low resistance in series

Q18. Emf of cell is measured in

1.     joule         (b)  joule/coulomb         (c)  joule-coulomb        (d)    joule/coulomb/meter

Q19. Potentiometer measures the potential difference more accurately than a voltmeter because 

     (a)   It has a wire of high resistance                            (b)     It has a wire of low resistance 

     (c)   It does not draw current from external circuit (d)     it draws heavy current from external

Q20. Copper wire is used as connecting wire because

1.    copper has high electrical resistance            (b) copper has low electrical resistance

(c)  Copper has low electrical conductivity     (d) copper has high value at elasticity

Q21. Ohm’s law is valid when temperature if conductor is

      (a)   Very low       (b)   very high     (c)   varying        (d) constant

Q22. While measuring emf of cell potentiometer acts as ideal voltmeter of

1.         high resistance      (b) low resistance       (c) infinite resistance      (d) zero resistance

Q23. Constantan wire is used for making standard resistance, because it has

    (a)  High melting point                     (b) low specific resistance 

    (c) High specific resistance               (d) negligible temperature coefficient of resistance

Q24. Appliances based on heating effect of current work on

     (a) Only a.c.                        (b) Only d.c.    (c) Both a.c. and d.c.            (d) None of these

Q25. Potentiometer is based on

     (a) Deflection method    (b) zero deflection method      (c)    both (a) and (b)        (d) None of these

Directions: Read the following questions and choose

 (A)  If both the statements are true and statement-2 is the correct explanation of statement-1. 

 (B) If both the statements are true but statement-2 is not the correct explanation of statement-1.

 (C) If statement-1 is True and statement-2 is False.

 (D)  If statement-1 is False and statement-2 is True.

1. Statement-1:  There is no current in the metals in the absence of electric field. 

 Statement-2:  Motion of free electrons are random. 

 (a) (A)  (b) (B) (c) (C) (d) (D)

2. Statement-1:  In meter bridge experiment, a high resistance is always connected in series with a galvanometer. 

 Statement-2:  As resistance increases current through the circuit increases.

 (a) (A)  (b) (B) (c) (C) (d) (D)

3. Statement-1:  A potentiometer of longer length is used for accurate measurement. 

 Statement-2:  The potential gradient for a potentiometer of longer length with a given source of emf becomes small.

 (a) (A)  (b) (B) (c) (C) (d) (D)

4. Statement-1:  Potential difference across the battery is always equal to emf of the battery. 

 Statement-2:  Work done by the battery per unit charge is the emf of the battery.

 (a) A (b) B (c) C (d) D

5. Statement-1: In a simple battery circuit the point at the lowest potential is positive terminal of the battery. 

 Statement-2: The current flows towards the point of the lower potential in the circuit, but it does not flow in a cell from positive to the negative terminal. 

 (a) (A)  (b) (B) (c) (C) (d) (D)

IMPORTANT QUESTION:- 

1. STATE AND EXPLAIN DRIFT VELOCITY………………..(3)
2. EXPLAIN OHM’S LAW………………………………………………(2)
3. DERIVE EXPRESSION FOR INTERNAL RESISTANCE OF A CELL………(3)
4. EXPLAIN METRE BRIDGE OR SLIDE WIRE BRIDGE……………..(5)
5. GIVE THE FACTOR ON WHICH RESISTANCE OF A CONDUCTOR DEPENDS…..(2)
6. GIVE DIFFERENCE BETWEEN EMF AND TERMINAL POTENTIAL DIFFERENCE…(2)

19. Potentiometer:- ^m.imp

A potentiometer is a device which is used to measure a unknown emf or potential difference accurately.

Construction:-

A potentiometer is consist of a long wire AB of length four metre fixed on a wooden board with the help of a copper strips along with metre scale. The ends A & B are connected to a strong battery, key and rheostat. Here rheostat provides a constant current, when Jockey is moved on the wire.

Principle:-

The potentiometer is based on the principle that when a constant current flow across the length of the wire of uniform area, then the potential drop is directly proportional to length of the wire.

Working:-

As shown in the fig:- If we connect a voltmeter between the end A & the Jockey, then we can see that when jockey move on the wire then potential difference varies in the voltmeter as 

Or  ρ(R= ρ)

Here  Constant called potential gradient 

So

O r

This is the principle of the potentiometer.

• The potentiometer has the advantage that it draws no current from the voltage source being measured.
• As such it is unaffected by the internal resistance of the source.
• In general alloys like constantan or manganin are used as potentiometer wire.
  Constantan or manganin are used as potentiometer wire because of the following two reasons which are listed below:-

Constantan or manganin wire posses high specific resistance

Constantan or manganin wire posses low temperature coefficient.

Q.17 how a potentiometer behave as an ideal voltmeter.

A ideal voltmeter does not change the potential difference of the circuit, it have infinite resistance. But it is not possible that a voltmeter have infinite resistance. But a potentiometer does not draw any current from the circuit, so it behaves as an ideal voltmeter.

20. Application of a potentiometer:- ^m.imp

1.       Comparison of emf of two primary cells.

A circuit diagram for comparing of emf of two cells is as shown in fig. in this diagram, the ends of the wire PQ are connected with a battery, a key & rheostat for obtaining constant current supply galvanometer is connected with jockey & the other end of the Galvanometer may be connected to a cell either 1 or 2 through a two way key system. When the connection 1 & 3 are joined & the jockey is moved at the wire then supposes at L1 the deflection in the Galvanometer becomes zero. At that situation:

— (I)

Now when the connection 1 & 3 are joined then we obtain 

— (2)

Dividing eq (1) by (2) we get 

Or   

Hence we can compare emf of two cells by using potentiometer.

Q.18 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 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.

The balance condition of the potentiometer is given as

⇒   =2.25V

(b)To determine the internal resistance of a cell:-

As shown in fig. firstly connects the key K so that the constant current flows in the circuit with the help of a rheostat.In this situation when jockey is moved on the wire then suppose at AJ = L₁ the galvanometer shows no deflection. Now the emf of the cell E= potential difference across the length of the wire

Or   — (1)

Now when key K₁ is closed, then galvanometer shows no deflection at

  .

Then Potential difference between two poles of the cell

V= potential difference across the length  of the wire

I.e.  — (2)

Diving eqⁿ (1) by (2) we get

— (3)

We know that

) R

Using equation 3

Hence we can calculate internal resistance of the cell.

C. Determination of Potential difference using potentiometer:-

A battery of emf E is connected with resistance box R & a key between terminals A & B.A resistance R₁ is connected to cell E₁ and key K₁ in series.

Theory and working:-

After closing the key K₁ the current flows through resistance R₁ & when jockey is moved on the wire

then suppose at AJ =  ₁ the Galvanometer shows no deflection. 

Then the maximum amount of the current flowing in the circuit is 

I = 

So the potential drop across the length of the wire will be 

) R

Where R is the resistance of the wire of length L .

Now potential gradient = fall of potential per unit length

I,ek==( )

As

SoV =( ) 

SENSITIVENESS OF POTENTIOMETER

A potentiometer is sensitive if

(1)It is measure a very small potential difference and

(2) If there is large change in length for small change in potential difference.

The sensitivity of a potentiometer may be increased by increasing the length of the wire.

21 Difference between potentiometer & voltmeter 

Q.19 Figure 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 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) emf of the cell, E₁ = 1.02 V Balance point, 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,

(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. 

Q.20 Figure 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 resistor, R = 10.0 Ω

Balance point l₁ = 58.3 cm Current in the potentiometer wire = i

Hence, potential drop across R,  

Resistance of the unknown resistor = X

Balance point for this resistor, l₂ = 68.5 cm

Hence, potential drop across X,

E2 = iX  As 

so

or  x=

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. 

Q.21 Figure 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 resistance 9.5 Ωis 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, 

r =  =1.68Ω

Therefore, the internal resistance of the cell is 1.68Ω.1