SOLUTION OF ELECTRICITY

1. Define charge. Give its S.I unit.
   Ans. Electric charge is a basic physical property of matter that causes it to experience a force of attraction or repulsion when placed in an electromagnetic field. S.I unit of charge is coulomb denoted by C.
2. Define one ohm.
   Ans. If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 Ω. That is, 1 ohm = 1 volt x1 ampere.
3. State Coulomb’s law.
   Ans. The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
4. What do you mean by the statement “Potential difference between points A and B in an electric field is 3 volts”?
   Ans. 3 volts is the potential difference between two points in a current-carrying conductor when 3 joules of work is done to move a charge of 1 coulomb from one point to the other.
5. Give the relation between 1 volt, 1 joule, and 1 coulomb. Name the physical quantity.
   Ans. The relation between 1 volt, 1 joule, and 1 coulomb is:
   1V = 1J / 1C.
6. The length of a wire is doubled. How is its resistivity affected?
   Ans. The length of a wire is doubled then the resistivity will remain the same because resistivity is the property of conductors.
7. A current of 1 A is drawn by a filament of an electric bulb. What would be the number of an electron passing through a cross-section of the filament in 16 s?
   Ans. If charge is Q, flows across any cross-section of a conductor in time t, then the current I, through the cross-section is 
   I = Q / t
   Here I = 1 A, t = 16s 
   Q = Ixt = 1 x 16 = 16C
   Now 1 C = 6.25 x 10¹⁸ electrons.
   Therefore 16 C = 16 x 6.25 x 10¹⁸ electrons. 
   = 10²⁰ electrons. 
8. A cylindrical conductor of length 'L' and uniform area of cross-section 'A' has resistance 'R'. What is the area of cross-section of another conductor of length '2L' and resistance 'R' of the same material? 
   Ans. According to the expression of resistance 
   R = ρ  L 
             A 
   or AR = ρL,  
   Crosssection area A is directly proportional to the length L so the area of cross-section of another conductor of length '2L will be double means 2A.
9. What is the maximum resistance which can be made by using five resistors each of 1/5 Ω?
   Ans. Maximum resistance can be obtained when the resistors are connected in series. R = R₁+ R₂+ R₃+ R₄ + R₅.
   R = 1/5 +1/5 +1/5 +1/5+ 1/5 = 5/5= 1Ω .
10. What is the minimum resistance which can be made by using five resistors each of 1/5 Ω?
    Ans. Minimum resistance can be obtained when the resistors are connected in parallel.
    Given each resistance R = 1/5 Therefore 1/R = 5 Ω
    1/R = 1/R₁+ 1/R₂+1/R₃+1/R₄++ 1/R₅.  
    1/R = 5 + 5 + 5 + 5+ 5 =25Ω.
    1/R = 25 Ω  Therefore R = 1/25 Ω .
11. What is the minimum resistance which can be made by using 2 Ω, 3 Ω and 5 Ω?
    Ans. Minimum resistance can be obtained when the resistors are connected in parallel.
    Given each resistance R₁ = 2Ω, R₂ = 3Ω and R₃ = 5Ω  
    1/R = 1/R₁+ 1/R₂+1/R₃+1/R₄++ 1/R₅.  
    1/R = 1/2 +1/3 + 1/5 = 15/30 + 10/30 + 6/30 = 31/30 Ω.
     1/R = 31/30 Ω , Therefore R = 30/31 = 0.97Ω.
12. Which is having more resistance, (a)  50 W bulb or (b)  25 W bulb?
    Ans. We have P = V²/R, If V is constant then R is inversely proportional to P.
    Therefore 25 W bulb has more resistance.
13. Write an expression for voltage (V).
    Ans. The expression for Voltage V = IR, Where R = resistance and I = Current(A).
14. If the current I through a resistor is increased by 100% at a constant temperature, what will be the increase in power dissipated?
    
    Ans. We have P = I²R, If R is constant then P is directly proportional to I².
    If the current I through a resistor is increased by 100% means doubled (I+I), then increase in power Pn = (2I)²R = 4I²R = 4P,
    Change in power
    Therefore the increase in power dissipated will be300%. 
15. Write the relation between resistance (R) of the filament of a bulb its power (P) and a constant voltage (V) applied across it.
16. Ans. The relation between resistance (R) of the filament of a bulb its power (P) and a constant voltage (V) is P = V²/R.
17. The voltage-current (V-I) graph of a metallic conductor at two different temperatures T₁ and T₂ is shown in Fig 1. At what temperature is the resistance higher? 
    
    Ans. We have V= IR, and R is directly proportional to temperature. Since T₁ has more slope than T₂, thus resistance is higher at T₁. 
18. Name the physical quantity whose unit is volt/ampere.
    Ans. The physical quantity is R (resistance) whose unit is volt/ampere.
    R = V/I.
19. Calculate the number of electrons constituting one coulomb of charge. Charge on 1 electron =1.6 ×10^-19 .
    Ans. We know that an electron possesses a negative charge of 1.6 × 10^-19 C.
    We have Q = ne, where n = no. of electrons, Q = Charge, and
    e = charge on each electron = 1.6 × 10^-19 C
    Now n = Q/e
    n = 1/1.6 × 10^-19
    n= 0.625 x 10¹⁹
    No. of electrons n = 6.25 x 10¹⁸.
    6.25 x 10¹⁸ electrons.
20. A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire changes?
    Ans. The resistance of a wire is directly proportional to the length of the wire. As its length increases it's resistance increases. According to the question, the length of the wire becomes 4 times. Therefore if the length is L then-new length will be 4L, and its resistance becomes 4x.
21. 400 J of heat is produced in 4 s.in a 4Ω resistor. Find potential difference across the resistors.
    Ans. Given H = 400J, t = 4s, and R = 4Ω
    We have H = I²Rt
    or H = (V/R)²R t = Rt V²/R²  = t V²/R
    Therefore V² = HR/t = (400 x 4)/4= 400 
    V = 20 v.
22. Which is having more resistance R (a) - 220 V,100 W bulb or (b) 220 V, 60 W bulb?
    Ans. We have P=V²/R, If V is constant then R is inversely proportional to P.
    Therefore (b) 220 V, 60 W bulb has more resistance. 
23. In a circuit, if two resistors of 5 Ω and 10 Ω are connected in series, compare the current passing through the two resistors?
    Ans. The current passing through the two resistors will be the same as both the resistors are in series.
24. What is the commercial unit of energy?
    Ans. The commercial unit of electric energy is a kilowatt-hour (kW h).
    1 kW h = 1000 watt × 3600 second = 3.6 × 10⁶ watt-second = 3.6 × 10⁶ joules (J).
25. Name the instrument used for the measure in     
    (a) Potential difference  (b) Current.
    Ans. The instrument used for the measure :
    (a) Potential difference- Voltameter
    (b) Current - Ammeter.
26. How is an ammeter connected in a circuit to measure the current flowing through it?
    Ans. It is always connected in series in a circuit through which the current is to be measured.
27. What happened to the resistance of a conductor when its area of cross-section is increased?
    Ans. According to the expression of resistance, 
    R = ρ  L 
              A 
    Resistance R is inversely proportional to the area of cross-section A, so if the area of cross-section of the conductor has increased the resistance will be decreased.
28. Through which of the two wires of the electric current will flow more easily:  
    (a) a thick wire or  
    (b) a thin wire,
    of the same material and of the same length when connected to the same source.
    Ans. The electric current will flow more easily through a thick wire as the current is inversely proportional to the resistance and resistance is inversely proportional to the thickness of the conductor. A thick wire has less resistance than thin, so the current flow through thick wire more easily.
29. What is electric power? State its SI unit.
    Ans. The rate of consumption of energy in an electric circuit is electric power. The power P is given by P = VI Or P = I²R = V²/R
    The SI unit of electric power is the watt (W).
30. State in brief the meaning of an electric circuit.
    Ans. An electric circuit is a continuous and closed path of an electric current.
31. Keeping the potential difference constant the resistance of an electric circuit is doubled. State the change in the reading of an ammeter connected in the circuit.
    Ans. The current in a circuit is inversely proportional to the resistance of the wire. As the resistance of an electric circuit is doubled, the flow of current decreases and becomes half. 
32. Draw a simple circuit diagram that can be used to verify Ohm’s law.
    Ans. According to Ohm's law: At the constant temperature, the current passing through a conductor is directly proportional to the potential difference applied across. V=IR
    
33. Define electricity.
    Ans. Electricity is the number of charges flowing through a particular area in unit time or it is the rate of flow of electric charges. S.I unit is ampere A.
34. What does an electric circuit mean? 
    Ans. A continuous and closed path of an electric current is called an electric circuit.
35. Define one ampere.
    Ans. One ampere is constituted by the flow of one coulomb of charge per second, that is, 1 A = 1 C/1 s. 
36. What is an ammeter?
    Ans. An instrument called ammeter measures the magnitude of electric current flowing in a circuit. It is always connected in series in a circuit through which the current is to be measured.
37. Name a device that helps to maintain a potential difference across a conductor.
    Ans. An instrument called voltmeter measures the potential difference between the two points of a circuit. It is always connected in parallel in a circuit. It has very high resistance. It is denoted by V.
38. What is a galvanometer?
    Ans. A galvanometer is a device is used to detect the direction of current or the presence of very weak current in the circuit. It is denoted by G.
39. What are the units of electric current? 
    Ans. The unit of electric current is ampere A.  Small quantities of current are expressed in milli-ampere (1 mA = 10^-3 A) or in micro-ampere (1 µA = 10^-6 A). 
40. A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
    Ans. We are given, I = 0.5 A; t = 10 min = 600 s.
    From Eq, we have Q = It = 0.5 A × 600 s = 300 C.
41. 10. What do you mean by the statement “Potential difference between points A and B in an electric field is 3 volt”?
    Ans. 3 volt is the potential difference between two points in a current-carrying conductor when 3 joules of work is done to move a charge of 1 coulomb from one point to the other.
42. How much energy is given to each coulomb of charge passing through a 6 V battery?
    Ans. We know that Work done = Potential difference x Charge
    Given: Potential difference = 6V,  Q = 1 C
    Work done= VQ= 6 x 1 = 6 J.
43. Define one ohm.
    Ans. If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 Ω. That is, 1 ohm = 1 volt x1 ampere.
44. What is A variable resistance?
    Ans. A component used to regulate current without changing the voltage source is called variable resistance. In an electric circuit, a device called rheostat is often used to change the resistance in the circuit. 
45. What happens to the current, If the resistance is doubled? 
    Ans. The current through a resistor is inversely proportional to its resistance. If the resistance is doubled the current gets halved.
46. On what factors does the resistance of a conductor depend?
    Ans. The resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material.
47. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 sec.Ans. Heat developed in electric iron H = I²Rt
    H =  5² x 20 x 30 = 25 x 20 x 30 = 15000 J.
48. Why does the cord of an electric heater not glow while the heating element does?
    Ans. According to the expression of heat H = I²Rt heat, produce is directly proportional to resistance and the resistance of the cord of an electric heater is very small so no heat is produced.
    
    The resistance of the heating element is very large so it gets red hot and glows.
49. What determines the rate at which energy is delivered by a current? 
    Ans. The rate of consumption of electric energy in an electric appliance is called electric power.P=IV
    
    Power is the rate at which energy is delivered by a current.
50. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V. 
    Ans. Given Q = 96000 C, potential difference  = 50 V
    We have H =I²Rt = I x V/R X R X t = IVt = Q/t X V x t =QV
    
    H = QV = 96000 x 50 = 48 x10⁵J= 4.8 x 10⁶J.
51. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
    Ans. Given R = 20 Ω, I = 5A and t = 30s
    
    H = =I²Rt = 25 x 20 x 30 = 15000J = 1.5 x 10⁴J.
52. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
    Ans. Given I =5A , V= 220 and t = 2hr
    We have P = IV
    P = 5 x 220 = 1100 W = 1.1 kWh
    
    Energy consumed = P x t = 1.1 x 2 = 2.2J.
53. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –
     (a) 1/25 (b) 1/5 (c) 5 (d) 25 
    Ans. R′ = 1/R' = 5/R + 5/R + 5/R + 5/R +5/R = 5 x 5/R
    1/R' = 25/R
    R/R' = 25/1 = 25
    Ans is (d).
54. Which of the following terms does not represent electrical power in a circuit? 
    (a) I² R  (b) I R² (c) VI (d) V²/R 
    Ans. (b) I R²
    
    I R²= I x R x R = VR  doesn't equal to P.
55. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
     (a) 100 W (b) 75 W (c) 50 W (d) 25 W .
    Ans. (c) 50 W
    P= IV  and I = P/V = 100/220 = 50W/110V 
    
    50W power will be consumed when it is operated on 110 V.
56. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶Ω, (b) 1 Ω and 10³ Ω, and 10⁶ Ω.
    
    Ans. (a) The equivalent resistance in case of (a) and (b) is less than 1 because in parallel equivalent resistance is less than individual resistance.
57. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. 
    Ans. Given potential difference = 12V, Current I = 2.5 mA
     = 2.5/1000 = 0.0025A.
    
    R = V/I = 12/0.0025 = 120000/25 = 4800 Ω.
58. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? 
    Ans. Given potential difference = 9V ,
    Total resistance R = 0.2 + 0.3 +0.4 + 0.5 +12 = 13.4 Ω
    I = V/R = 9/13.4 = 90/134 =0.67A 
    0.67 A current will flow through the 12 Ω resistor because in series same current flow.
59. What determines the rate at which energy is delivered by a current? 
    Ans. The rate at which energy is delivered by a current is power. It is denoted by P.

II. Short Answer Type Questions: 

1. State  Ohm’s law
   
   Therefor V ∝ I or V/I = constant = R or V = IR 
   
   resistance
   Ans. According to German physicist George Simon Ohm, the electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains the same. This is called Ohm’s law.
2. How much work is done in moving a charge of 2 C across two points having a potential difference 12 V? 
   Ans. The amount of charge Q, that flows between two points at potential difference V (= 12 V) is 2 C. 
   Thus, the amount of work W, done in moving the charge is 
   W = VQ = 12 V × 2 C = 24 J.
3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
   
   
   Ans.  If the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value, the current gets half because current is directly proportional to the potential difference. 
   We have I = V/R.
   If V' =1/2V 
   I' = V'R = V/2R =1/2 I. 
   The current gets half. 
4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
   
   Ans.  The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize or burn readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc.
5. (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? 
   Ans.  (a) We are given V = 220 V; R = 1200 Ω. We have the current
    I = V/R = 220 V/1200 Ω = 0.18 A. 
   (b) We are given, V = 220 V, R = 100 Ω. We have the current 
   
   
   I = 220 V/100 Ω = 2.2 A. 
6. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V? 
   Ans. We are given, potential difference V = 60 V, current I = 4 A. 

   According to Ohm’s law, R = V/I = 60 V/4A = 15 Ω

   When the potential difference is increased to 120 V the current is given by current I= 120 V /15R = 8A. 

   The current through the heater becomes 8 A.
7. A hot plate connected to a 220 V line has two resistance coils A and B, each of 22-ohm resistance. Calculate the amount of electric current flowing when this coil is:
   
   
   (i) Used individually (ii) Connected in series  (iii) Connected in parallel.
   Ans. (i) When used individually :
   Given V = 220v, R = 22 Ω
   I = V/R = 220 / 22 = 10 A.
   (ii) When connected in series
   R = R₁ + R₂ = 22 + 22 = 44 Ω
   I = V/R = 220 / 44 = 5A.
   (iii) When connected in parallel. 
   1/R = 1/R₁ + 1/R₂ = 1/22 +1/22 = 2/22 = 1/11
   R = 11 Ω.
   I = V/R = 220/11 = 20 A.
8. A torch bulb is rated 5 volts and 500 mA. Calculate its

   (i) power (ii) resistance (iii) energy consumed when it is lighted for 4 hours.
   Ans. (i) Given V = 5v and I = 500mA = 500/1000 = 0.5 A
   Power P = VI = 5 x 0.5 = 2.5 W.
   (ii) R = V/I = 5/0.5 = 50/5 = 10Ω.
   (iii) Energy consumed W = P x t = 2.5 x 4 x 60 x 60 =36000 J =3.6 x 10⁴J .
9. Which consumes more energy a 250 watt TV set in 1 hour or 1200W toaster in 10 min?
   Ans. Energy consumed by TV set = P  x t = 250 w x 1h = 250 wh.
   Energy consumed by toaster = p x t = 1200w x 10min
   Energy consumed = 1200 x 10/60 = 1200/6 = 200 wh.
   
   TV set consumes more energy.
10. (a) Define the electric power of a device of resistance R is connected across a source of  V voltage and draws a current I. Derive an expression for power in terms of voltage and resistance.
    (b) An electric bulb is connected to a 220 V generator. The current is 0.5 A. What is the power of the bulb?
    Ans. (a) The rate at which electric energy is dissipated in an electric circuit is called power.
    P = V.Q/t 
    = VI 
    = V x V/R = V²/R
    P = V²/R
    
    (b) Given V = 220V,  I = 0.5A 
    
    P = VI = 220 x 0.5 = 110 W.
11. Explain the following:
    (i) Series arrangement is not used for domestic circuits.
    (ii) Copper and Aluminum wires are usually employed for electricity transmission.
    Ans. (i) In a series circuit, the current is constant throughout the electric circuit which is not required and when one component fails, the circuit is broken and none of the components works. So the series arrangement is not used for domestic circuits.
    
    (ii)  Copper and Aluminum wires are the best good conductor and have low resistivity.
12. What are the current flowing and potential difference across 10Ω resistance?

    

    Ans. In series combination, the same current flows. 

    V = 30 V  

    R = R₁+ R₂ + R₃ = 5Ω +10Ω + 15Ω = 30Ω

    I = V/R = 30/30 = 1A.
13. A lamp rated 60 W and an electric iron rated 800 W are used for 6 hours every day. Calculate the total energy consumed in 30 days.
    Ans. Energy consumed by lamp = P x t = 60 W x 6 hrs x 30 days
    Energy consumed by electric iron  = P x t = 800 W x 6 hrs x 30 days
    Total energy consumed 
    = 6 x 30 (60 + 800) 
    = 180 x 860 
    = 154800 Whr
    
    E = 154.800 kWh.
14. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? 
    Ans. Resultant resistance R = Resistance of electric iron.
    1/Ri = 1/100 + 1/50 + 1/500
    1/Ri= (5+10+1)/500
    1/Ri=16/500
    Ri = 31.25 Ω
    V= 220 v
    
    I = V/R = 220/31.25 =7.04 A.
15. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series? 
    Ans. The equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
    1/R= 1/R₁ + 1/R₂ + 1/R₃.
    
    Therefore total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has a different resistance and requires different currents to operate properly.
16. A number of bulbs are to be connected to a single source. Will they provide more illumination when connected in parallel or in series? Give reasons to justify your answer.
    Ans. In parallel combination, bulbs are glowing brightly, because the same voltage is applied to each appliance and the voltage is not eaten up by the other appliances. 
    
    In a series connection, the main voltage is eaten up by each appliance due to which appliances would have low voltage and not glow brightly. 
17. What is (a) the highest (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

    Ans. (a) To get the highest resistance, the resistor should be connected in series.

    Therefor resultant resistance R = 4 + 8 + 12 + 24 = 48 Ω

    (b) To get the lowest resistance, resistor should be connect in parallel.

    Therefor Resultant resistance 

    1/R= 1/4 +1/8 + 1/12 +1/24 = 12/24 = 1/2

    R = 2 Ω.
18. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be -
     (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 .
    Ans. (d)
    Let there is a number of a resistor of resistance R.
    Rs = 2R
    Rp = R/2
    Ratio Rs/Rp = 2R     = 2²:1 = 4:1
                         R/2
    Hs    =  I² .Rs. t  = Rs/Rp = 4:1                [ H = I²Rt ]
    
    Hp        I² .Rp. t
19. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? 

    Ans. Let the number of resistance be n.

    1/Rp = n/R 

    Rp = R/n

    Given I = 5A , R= 176 Ω and V=220v

    Rp = V/I 

    Rp = 220/5 = 44

    n = R/Rp= 176/44 = 4 resistors.
20. The current is different for different components. Why? 
    
    
    Ans. Certain components offer an easy path for the flow of electric current while others resist the flow. The motion of electrons in an electric circuit constitutes an electric current. The electrons, however, are not completely free to move within a conductor. They are restrained by the attraction of the atoms among which they move. Components that offer low resistance is a good conductor. A component that offers a higher resistance is a poor conductor. An insulator offers higher resistance.
21. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why? 
    Ans. The current through a resistor is inversely proportional to its resistance and resistance of the conductor is inversely proportional to its area of cross-section. Thus thick wire has low resistance and current will flow more easily through a thick wire.
22. Answer the following:
    (a) Which among iron and mercury is a better conductor? 
    (b) Which material is the best conductor?
    Ans. (a) The resistivity of iron is 10.0 × 10^-8 and the resistivity of mercury is 94.0 × 10–8.
     The resistivity of mercury is greater than iron so iron is a better conductor.
    (b) Silver is the best conductor because it has at least resistivity.
23. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
    
    Ans. Given : Resistance R  = 8 Ω
    Current I = 15 A
    We know that heat energy produced in heater H = I²Rt
    the rate at which heat is developed in the heater 
    P = H/t = I²R = 15×15×8 = 1800 w or J/s
    Heat is produced by the heater at the rate of  1800 J/s. 
24. What is meant by saying that the potential difference between two points is 1 V? 
    
    
    Ans. One volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. Therefore, 1 volt = 1 joule 1 coulomb. 1 V = 1 J / C.
25. Define Potential difference. What is its S.I unit?
    Ans. The electric potential difference between two points in an electric circuit is equal to the work done to move a unit charge from one point to the other in a conductor.
    Potential difference (V) = Work done 
    =(W)/Charge (Q). S.I unit of V is J/C.
26. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V? 
    Ans. We have potential difference V = 220 V
    Current drawn by the bulb of rating 100 W:
    

    I=P/V

     =100/220=0.45 A

    Similarly, current drawn by the bulb of rating 60 W 

    I=P/V

     =60/220=0.27A

    The total current drawn from the line I =0.45 + 0.27=0.72 A.
    
    

III. Long Answer Type Questions:

1. State resistivity.
   Ans. The resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A). 
   That is, R ∝ l and R ∝ 1/A 
    Combining Eqs. we get R ∝ l /A 
   or, R = ρ  l 
                  A 
   If the conductor is a cylinder, then A =πr², Where r is the radius of cross-section of the cylinder.
   R = ρ  l 
            πr² 
   where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is Ωm. It is a characteristic property of the material. 
2. The resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? 
   Ans. We are given the resistance R of the wire = 26 Ω, the diameter d = 0.3 mm = 3 × 10^-4 m, and the length l of the wire = 1 m.
   Therefore,  the resistivity of the given metallic wire is ρ = (RA/l) = (Rπd2/4l) 
   Substitution of values in this gives
    ρ = 1.84 × 10^-8 Ω m.
   The resistivity of the metal at 20°C is
    1.84 × 10^-8 Ω m. 
   This is the resistivity of manganese.
3. A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.
   
   Ans. We are given, R = 4 Ω. 

   When a wire is doubled on it, its length would become half and the area of cross-section would double. That is, a wire of length l and area of cross-section A becomes of length l/2 and area of cross-section 2A.  we have R = ρ l/A

   R1 = ρ l/2

             2A

   Where R1 is the new resistance.

   Therefore, R₁ / R= 1/4

   R₁ = R/4 = 4/4 = 1Ω

   The new resistance of the wire is 1Ω.
   
   
4. Derive the relation for equivalent resistance when three resistances are connected parallel.
   
   Ans. We know that the total current is equal to the sum of the currents through each branch of the combination I = I₁ + I₂ +I₃
   Image 
   Let Rp be the equivalent resistance of the parallel combination of resistors. By applying Ohm’s law to the parallel combination of resistors, 
   we have I = V/Rp  
   On applying Ohm’s law to each resistor, 
   we have I₁ = V /R₁ ;
    I₂ = V /R₂; and
    I₃ = V /R₃ 
   Therefor  V/Rp = V/R₁ + V/R₂ + V/R₃ or 
    1/Rp = 1/R₁ + 1/R₂ + 1/R3
   Thus, we may conclude that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
5. Derive the relation for equivalent resistance when three resistances are connected series.
   
   Ans.The potential difference V is equal to the sum of potential differences V₁, V_2, and V₃.That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is, V = V₁ + V₂ + V₃ 
   Image 
   let 'I' be the current through the circuit. The current through each resistor is also I. It is possible to replace the three resistors joined in series by an equivalent single resistor of resistance R, such that the potential difference V across it, and the current I through the circuit remains the same.
    Applying the Ohm’s law to the entire circuit, we have V = I R
   On applying Ohm’s law to the three resistors separately, we further have 
   V₁ = I R₁, V₁  = I R₂  and V₃  = I R₃  
   I R = I R₁ + I R₂ + I R3 
   or Rs = R₁ + R₂ + R₃.
   We can conclude that when several resistors are joined in series, the resistance of the combination Rs equals the sum of their individual resistances, R₁, R₂
   and R₃, and is thus greater than any individual resistance.
6. (a) Nichrome wire of length l and radius r has a resistance of 10Ω. How would the resistance of the wire change when:
   
   
   (i) Only the length of the wire is doubled?
   (ii) Only the diameter of the wire is doubled. Justify your answer?
   (b) Why elements of electrical heating devices are made up of alloys.
   Ans. (i) Resistance R is directly proportional to length L. 
   R ∝ L, So the resistance will be doubled when the length of the wire is doubled.
   i.e R = 2 x 10 = 20Ω
   (ii) Resistance R is inversely proportional to area A
   i.e R ∝ 1/A ∝ 1/d² 
   When the diameter of the wire  is doubled R ∝ 1/(2d)²∝1/4d²
   R becomes one forth.d
   (b) The element of electrical heating devices are made up of alloys because
   i. they have high resistance due to which they get heated up enormously.
   ii. they do not oxidize easily and
   iii they have a high melting point due to which they don't melt at high temperature.
7. How would you connect three resistors each of resistance 6Ω so that the combination has a resistance of (i) 9Ω (ii) 4Ω? Justify your answer.
   
   Ans. (i) 9Ω > 6Ω, so the combination should be in series but 6Ω + 6Ω = 12Ω which is greater than 9Ω, so at first there should be a parallel combination.
   1/R = 1/R₁ + 1/R₁ 
   1/R = 2/R₁
   1/R = 2/6 =1/3 
   R = 3Ω
   To get 9Ω, we have to connect another resistor of resistance 6Ω in series combination.
   Now total resistance R = 3Ω + 6Ω = 9Ω.
   (ii) 4Ω < 6Ω 
   At first there should be a series combination. R = 6Ω + 6Ω = 12Ω
   and then parallel combination. R = 1/12 + 1/6 = 3/12 =1/4 
   R = 4Ω.
8. A wire of resistance R is cut into 5 equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R'. Calculate the ratio R/R'. 

   Ans. a) The resistance of a wire is directly proportional to its length. 

   Let the resistance of wire be R. If the wire is cut into five equal parts

   then resistance of each part will be =R/5

   All the five parts are connected in parallel.

   equivalent resistance (R') is given as:

   1/R' = 5/R + 5/R + 5/R + 5/R +5/R = 5 x 5/R

   1/R' = 25/R

   R/R' = 25/1 = 25:1.
9. (a) State one difference between kilowatt and kilowatt-hour. Express 1 kWh in joules.
   
   (b) Calculate the amount of heat generated when 7200 coulombs of charge are transferred in one hour through a potential difference of 50 V.

   Ans.(a) Kilowatt is the unit of power and kilowatt hour is the unit of energy consumed by appliances. 
   1kWh = 1kW x hr = 1000 W x 3600 s = 3600000 J = 3.6 x 10⁶J
   
   (b) Here given Q = 7200 C, Potential difference = 50 V time t = 1hr =3600s
   Heat generated H = QV = 7200 x 50 = 360000 J = 3.6 x  10⁵J.
10. An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate 

    (a) the total resistance of the circuit, 
    (b) the current through the circuit, and 
    (c) the potential difference across the electric lamp and conductor. 
    Ans. (a) The resistance of electric lamp, R₁ = 20 Ω, 
    The resistance of the conductor connected in series, R₂ = 4 Ω. 
    Then the total resistance in the circuit R =R₁ + R₂ 
    R_S = 20 Ω + 4 Ω = 24 Ω.
    (b) The total potential difference across the two terminals of the battery.
    V = 6 V. Now by Ohm’s law, the current through the circuit is given by
    I = V/R_S = 6 V/24 Ω = 0.25 A.
    (c) Applying Ohm’s law to the electric lamp and conductor separately, we get potential difference across the electric lamp, V₁ = 20 Ω × 0.25 A = 5 V; and, that across the conductor, V₂ = 4 Ω × 0.25 A = 1 V.
11. An electric lamp is marked 20 W, 220 V. It is used for 10 hours daily. Calculate:
    
    (a) Its resistance while glowing.
    (b) No.of units consumed every day.
    Ans. (a) Given P = 20W and V = 220v
    We have P = VI = V x V/R   ( IR = V )
    P= V² /R 
    R = V²/P= 220 x 220
                          20
    R = 220 x 11 = 2420Ω.
    Resistance while glowing is 2420Ω
    (b) No. of units consumed every day E = P x t
     = 20W x 10hrs = 200Wh = 0.2kWh.
     No.of units consumed every day is 0.2kWh.
12. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω? 
    Ans. (a) Let R1 = 2 Ω  R2 = 3 Ω and R3 = 6 Ω.
    To get 4 Ω, we have to connect a resistor of resistance 3 Ω and 6 Ω in parallel combination and then connect in series combination with resistance 2 Ω.
    1/R = 1/3 + 1/6 = 3/6 Ω
    1/R = 1/2
    Resultant resistance R = 2 Ω
    Now after connecting in series combination with resistance 2 Ω
    Resultant resistance = 2 Ω + 2 Ω = 4 Ω
    (b) To get total resistance 1 Ω we have to connect all resistors in parallel combination.
    Total resistance 1/R = 1/2 Ω + 1/3 + 1/6 = 6/6 = 1 Ω.
13. cA copper wire has a diameter of 0.5 mm and a resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled? 
    Ans. Area of cross-section of the wire, A =π (d/2) ²
    

    Diameter= 0.5 mm = 0.0005 m, radius r = 0.0005/2 

    Resistance, R = 10 Ω

    We know that

    R = ρ l/A

    l = RA/ρ = R x π x r^2 

                       ρ     

    = 10 x 3.14 x 25 x 10^-8

        1.6 x 10^-8 x 4

    = 10 x 3.14 x 25

              1.6 x 4

    = 122.66 m
    

    length of the wire = 122.66 m

    If the diameter of the wire is doubled, new diameter=2×0.5=1 mm=0.001 m

    Let new resistance be Rʹ

    R' = ρ l/A

    R' = 1.6 x 10^-8 x 122.66 

              π(1/2 x 10^-3)²

    = 1.6 x 10^-8 x 122.66 x 4

            3.14 x  10^-6 

    = 250.2 x 10^-2

    = 2.5 

    The new resistance is 2.5 Ω .
14. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
    
    
    Ans.  Let the number of electric bulbs be n.
    Given P of each bulb = 10 W, Therefor total power of n bulb P = n 10 W
    V = 220 v and current I = 5 A
    Resultant resistance R'=R/n 
    P = IV 
    n 10 = 5 x 220 = 1100
    n = 110
    Number of lamps 110.
15. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases? 
    
    
    Ans. Potential difference V = 220 V
    The resistance of first coil R =24 Ω
    (i) When coils are connected separately:
    According to Ohm’s law, V=IR
    I=V/R=220/24=9.166 A
    The current flow through the coil when connected separately is 9.16 A.
    (ii) When coils are connected in series
    R=24 + 24 = 48 Ω
    According to Ohm’s law
    I=V/R=220/48=4.58 A
    Current flowing through the coil when connected in series is 4.58 A
    (iii) When coils are connected in parallel:
    According to Ohm’s law
    I=V/R=220/12=18.33A
    Current flowing through the coil when connected in  parallel is 18.33 A.
16. Compare the power used in the 2 Ω resistor in each of the following  circuits:
    
    (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
    (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors. 
    Ans. (i) Potential difference (V) = 6 V
    

     Here 1 Ω and 2 Ω resistors are connected in series.

    Resultant resistance of the circuit (R) = 1 + 2 = 3 Ω

    According to Ohm’s law

    V = IR

    Or I=V/R=6/3=2A

    the current through the circuit (I)=2A

    Now the power used in the 2 Ω resistor

    we know that Power = IV = I²R  [V=IR]

                                     = 2×2×2 = 8W

    (ii) Potential difference, V = 4 V

    12Ω and 2Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across the 2Ω resistor will be 4 V. Power consumed by 2 Ω resistor is given by

    P = IV = V^2/R = 16/2 =8  [ I=V/R]

    The power used by the 2 Ω resistor is 8 W.
17. Explain the following.
    
    
    (a) Why is the tungsten used almost exclusively for filament of electric lamps? 
    (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? 
    (c) Why is the series arrangement not used for domestic circuits? 
    (d) How does the resistance of a wire vary with its area of cross-section? 
    (e) Why are copper and aluminium wires usually employed for electricity transmission?
    

    Ans. (a) The melting point of tungsten is very and has very high resistivity so it does not burn easily at a high temperature.

    (b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because the resistivity of an alloy is more than that of metals, do not oxidize or burn readily at high temperatures and produce a large amount of heat.

    (c) In a series circuit, the current is constant throughout the electric circuit. Thus it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly and when one component fails the circuit is broken and none of the components works. Hence, a series arrangement is not used in domestic circuits.

    (d) The resistance of a wire is inversely proportional to its area of cross-section, i.e. when an area of cross-section increases the resistance decreases or vice versa.

    (e) Copper and aluminium are good conductors of electricity, have low resistivity, and produce very very less heat or negligible heat due to which very little electricity is wasted. So they are usually used for electricity transmission.