CBSE Class 10 Physics Electricity Worksheet Set C

ONE MARK QUESTIONS

Question: Why do we use copper and aluminium wire for transmission of electric current?
Answer: Copper and aluminium wires are used for electric transmission due to their low resistivity.

Question: List any two factors on which resistance of a conductor depends.
Answer: Resistance of a conductor:
a. is directly proportional to its length                    
                   
                    R ∝ p

b. is inversely proportional to its area of cross section.
                   
                   R∝ 1/A

Combining (1) and (2), we get
               
                   R∝p/A

Question: Mention one reason why tungsten is used for making filament of electric lamp.
Answer: Tungsten is used for making filament because of its high melting point and low resistivity.

Question: What is the SI unit of electric potential?
Answer: Volt is the SI unit of electric potential.

Question: Write SI unit of resistivity.
Answer: Ohm-m

Question: If the charge on an electron be 1.6 x10-¹⁹C, find the approximate number of electrons in 1 C.
Answer: 1.6×10-¹⁹C charge is of = 1 electron and
              1 C charge is of =1/1.6×10^-19 electron

              No. of electrons = 6.25×10¹⁸    

Question: In an electric circuit, state the relationship between the direction of conventional current and the direction of flow of electrons.
Answer: Electrons flows from negative terminal to positive terminal where as current flows from +ve terminal to -ve terminal in external circuit i.e. Conventional current and electrons flow are opposite to each other.

Question: Power of a lamp is 60 W. Find the energy in joules consumed by it in 1 s.
Answer: P = 60W, t = 1 s
Energy = (VI)t
E = Pxt = 60×1 J
E = 60 J

TWO MARKS QUESTIONS

Question: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hours. 
Answer: I = 5 A, V = 220 V, P = ?, E = ?, t = 2 hours
P = V × I = 220 V × 5 A = 1100 W
E = P × t = 1100 W × 2 h = 2200 W h = 2.2 kW h

Question:  A thick wire and a thin wire of the same material are successively connected to the same circuit to find their respective resistance. Which one will have lower resistance? Give reason.
Answer: As resistance ∝1/A i.e. more area of cross-section lesser the resistance and vice versa. So thick wire has lower resistance.

Question:  How much current will an electric bulb of resistance 1100 W draw from a 220 V source? If a heater of resistance 100 W is connected to the same source instead of the bulb, calculate the current drawn by the heater.
Answer: Resistance of bulb, R = 1100 Ω
                                       
                                         V = 220 volt
                                     
                                       V = IR or I= V/R
                                     
                                     I=220\ 1100=1/5A
         
           When heater is connected with the same source then

                        I=V/R=220/100=2.2A

Question:  State the factors on which the heat produced in a current carrying conductor depends. Give one practical application of this effect.
Answer: We know that H = VIt or H = I2Rt Heat produced in a current carrying conductor
H ∝ I² (Square of the current in the circuit)
∝ R (Resistance of the conductor)
∝ t (Time for which current is passed in conductor) This effect applied in electric heater.

Question: The amount of charge passing through a cell in four second is 12 C. Find the current supplied by cell.
Answer: Given: t = 4 s
Q = 12C
I t
Q = 4 A
= 12 = 3 A

Question: What is the cost of running an AC with average power of 1000 W for 8 hours for 30 days. The cost of electric energy is `4.70 per kW h.
Answer: E = P × t = 1000 W × 8 × 30 = 240000 W h = 240 kW h
Cost of electric energy = 240 × 4.70 = rs 1128.

Question: How are ammeters and voltmeters connected in a circuit? What do they help us measure?
Answer: An ammeter which measure the current in a circuit is connected in series. Voltmeter is used to measure potential difference across a conductor so it is connected in parallel to it.

 Question: An electric iron has a rating of 750 W, 220 V. Calculate the
(i) Current flowing through it and (ii) Its resistance when it is in use.
Answer: (i) P = 750 W, V = 220 V
P = V × I ⇒ 750 = 220 × I ⇒ I = 750/220
= 3.409 A

Question: Why do we get electric shock in damp conditions?
Answer: Water provides conducting path for a current to flow through the human body in damp conditions like bathroom. We get electric shock if we are bare footed. Wet body has low resistance, high current can easily pass through, leading to electric shock.

Question:  Name the device/instrument used to measure potential
difference. How is it connected in an electric circuit?
Answer: The device which is used to measure potential  difference is voltmeter. Voltmeter is connected in parallel in an electric circuit.

Question: An electric bulb draws a current of 0.2 A when the voltage is 220 volts. Calculate the amount of charge flowing through it in one hour.
Answer: Given: I = 0.2 A                      
                       V = 220 Volt                       
                       t = 1 hr.
                       Q = ?
                       I = Q/ t or Q =I X T
                      Q=0.2 A X 1 hr =0.2 X  60 X 60 A – s
                   =720 C   

Question: An electric iron of resistance 20 W takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:R = 20 Ω, I = 5 A, t = 30 s
H = I² × R × t = 5 × 5 × 20 × 30 = 15000 J = 15 kJ

Question: (a) What material is used in making the filament of an electric bulb?
(b) Name the characteristics which make it suitable for this.
Answer: a. Tungsten is used in making filament.
b. Its high resistivity and high melting point.

Question: An electric fan has a rating of 460 W on the 230 V mains line. What fuse should be fitted in the plug?
Answer: P = V × I ⇒ 460 = 230 × I ⇒ I = 2 Ampere
A 3A fuse should be fitted (It should be slightly more than 2A).

Question: An electric kettle of 2 kW works for 2 hours daily. Calculate the
(a) Energy consumed in SI unit and commercial unit.
(b) Cost of running it in the month of June at a rate of ` 3.00 per unit. 
Answer: (a) P = 2 kW, t = 2 h
E = P × t = 2 × 2 h = 4 kW h
(b) Total energy consumed per month = 4 kW h × 30 = 120 kW h
Total cost = 120 × 3 = rs 360

Question: Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. 
Answer: Q = 96000 C, t = 1 hr = 3600 s, V = 50 volt
H = V × I × t = V × Q = 50 × 96000 C =(50 × 96000)/1000 kJ = 4800 kJ

Question: (i) Explain what is the difference between direct current and alternating current. Write one important advantage of using alternating current.
(ii) An air conditioner of 2 kW is used in an electric circuit having a fuse rating of 10 A. If the potential difference of the supply is 220 V, will the fuse be able to withstand when the air conditioner is switched on? Justify your answer. 
Answer: (i) The current whose direction gets reversed after every half cycle is called alternating current or A.C.
There is no change in the direction of D.C. D.C. is a uni directional current.
The most important advantage of using A.C. over D.C. is that in the A.C. mode electric power can be transmitted over long distances at a high voltage and low current with very little loss of power.
(ii) Here P = 2 kW = 2000 W, V = 220 Volt
P = VI, so the current, I = P/V =2000/220 = 9.09 A
As the current is 9.09 A below the rating of fuse, therefore the fuse will withstand, i.e. it will not blow off when A.C. is on.

Question: What is meant by ‘electrical resistance’ of a conductor? State how resistance of a conductor is effected when
(i) Low current passes through it for a short duration.
(ii) A heavy current passes through it for 30 seconds. 
Answer: It is in opposite direction offered to the flow of electric current.
(i) No effect on resistance, low current, hence no appreciable rise in temperature so there no change in resistance.
(ii) Heavy current for 30 seconds may increase the temperature so resistance will increase.

THREE MARKS QUESTIONS

Question: (a) Why are copper or aluminium wires generally used for electrical transmission and distribution purposes?
(b) Two wires, one of copper and other of manganin, have equal lengths and equal resistances. Which wire is thicker? Given that resistivity of copper is lower than that of manganin.
Answer: a. Copper or aluminium wires are used for transmission and distribution of electricity due to their low resistivity and high conductivity.
b. We know that R=p l/A
                          p∝A

Thicker the wire, more the resistivity. The resistivity of manganin is more than copper. So manganin wire is thicker than copper.

Question: Find the number of electrons transferred between two points kept at a potential difference of 20 V if 40 J of work is done.
Answer: 
Given: V = 20 Volt          
W = 40 J           
W = P X t         
= V X I X t        
 =V X Q/t X t
or W = V X Q
40 = 20 X Q
or Q =2 C
n e = Q
n =Q/e=2/1.6 X 10¹⁹
1.25 X 10¹⁹

Question: What is meant by “electrical resistance” of a conductor? State how resistance of a conductor is affected when
a. a low current passes through it for a short duration;
b. a heavy current passes through it for about 30 seconds.
Answer: Electrical resistance is the property of a conductor by virtue of which it opposes the flow of current through it. It is equal to the ratio of the potential difference applied across its ends to the current flowing through it.
                    R= V/I 
a. When a low current is passed for a short duration, through a conductor, heat produced is almost negligible and hence no appreciable change in its resistance.
b. When heavy current is passed through the conductor for 30 s. Conductor may be get heated and its resistance and resistivity change.

Question:  (a) Define the term ‘volt’.
(b) State the relation between work, charge and potential difference for an electric circuit.
Calculate the potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.
Answer: a. Potential difference b/w two points in an electric field is said to be 1 volt if the amount of work done in bringing a unit positive charge from one point to another point is 1 J.
b. Given: W = 100 J, Q = 20 C, V = ?
As      V=W/ Q ⇒ V=100/20 J C^-1
V = 5 J C^-1

V = 5 Volt.

Question: Name and define the SI unit of current. Calculate the number of electrons that flow through a conductor in 1 second to constitute a current of 1 ampere. (Charge on an electron = 1.6 X 10-¹⁹ coulomb)
Answer: SI unit of current is Ampere (A)
                                   
                            I=q/t      
                                      
              If q = 1 C, t = 1 s
then           I = 1 A
If 1 C charge flows in 1 s in a conductor then magnitude of current is said to be 1 A.

q = n e                              
                                 
n= q/e = I X t/e
                               
=1 A X 1 s / 1.6 X 10-¹⁹=100/16 X 10¹⁸
                                 
=6.25 X 10¹⁸

FIVE MARKS QUESTIONS

Question: Two identical wires, one of nichrome and other of copper are connected in series and a current (i) is passed through them. State the change observed in the temperature of the two wires. Justify your answer. State the law which explains the above observations.
Answer: The resistivity of nichrome is more than that of copper, so its resistance is also high. Therefore large amount of heat is produced in the nichrome wire for the same current passed as compared to that of the copper wire. Accordingly greater change in temperature is observed in case of nichrome wire. This can be explained by Joule’s law of heating.
Joule’s law of heating states that the amount of heat produced in a conductor is:
(i) Directly proportional to the square of current flowing through it, i.e.
H ∝ I²
(ii) Directly proportional to the resistance offered by the conductor to the flow current, i.e.
H ∝ R
(iii) Directly proportional to the time for which current is flowing through the conductor, i.e.
H ∝ t
Combining these, we get, H ∝ I²Rt, or H = KI²Rt
where K is a proportionality constant and in SI units, it is equal to unity.

Question: Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit.
Calculate the resistance of 50 m length of wire of cross-sectional area 0.01 square mm and of resistivity 5X10-8 Ωm .
Answer: Resistance is the opposition offered in the path of flow of current by the atoms or molecules of a conductor. Factors affecting resistances:
a. length R∝ l
b. area of cross-section R∝1/A
c. nature of material.
Rheostat is used to change the current in the circuit without changing the voltage source.
Given:     l = 50 m
              A = 0.01 mm² = 0.01X10-⁶ m²
              P=5 X10⁸ ΩM
As           R=P l/A=5X10⁸X 50/0.01X10-⁶
              R=250Ω

Question:(a) Define potential difference between two points in a conductor.
(b) Name the instrument used to measure the potential difference in a circuit. How is it connected?
(c) A current of 2 A passes through a circuit for 1 minute. If potential difference between the terminals of the circuit is 3 V, what is the work done in transferring the charges?
Answer: a. Electric potential is the amount of work done in bringing a unit positive charge from one point to another.
b. Voltmeter. It is connected in parallel in the circuit.
c. I = 2A, t = 1 min. = 60 s, V = 3 V
W = VQ = V(It)
W = 3X2X60 J
W = 360 J

Question: (a) What is the function of fuse wire in an electric circuit?
(b) What would be the rating of the fuse for an electric kettle which is operated at 220 V and consumes 500 W power?
(c) How is the SI unit of electric energy related to its commercial unit?
Answer: (a) Fuse wire protects the electrical circuits from over voltage and high current.
(b) 2.2 A flows through the circuit, fuse should be rated 3 A.
(c) 1 kW h = 3.6 × 10⁶ J

Question: State Ohm’s law. Write the mathematical representation of Ohm’s law. Use this relationship to define 1 ohm. List two disadvantages of connecting different electrical appliances in series.
Answer: If the physical conditions of a conductor are kept constant then current is directly proportional to the potential difference applied across it.
Mathematical representation of Ohm’s law V = IR.
                       V/I=R
             I Volt/1 Ampere =1 ohm

If by applying a potential difference of 1 volt across a conductor, the current is 1A then the resistance of the conductor is said to be 1 ohm.
Two disadvantages of connecting electrical appliance in series.
a. If one appliance fails to operate then the circuit is broken and other devices also will not operate.
b. Different devices require different amount of current to operate but in series combination, same current is supplied to all electrical appliances.

 Question: (a) State the commercial unit of electric energy and find its relation with its SI unit.
(b) The current through a resistor is made three times its initial value. Calculate how it will affect the heat produced in the resistor.
(c) Find the increase in the amount of heat generated in a conductor if another conductor of double resistance is connected in the circuit keeping all other factors unchanged.
Answer: a. Commercial unit of electric energy = kWh
     1 kWh = 3. 6 X 1 0⁶ J
b. Initial heat generated in the
    resistor = I ² Rt
H ₁ = I ² Rt 
when current is made three times i.e. 3 I now heat generated
H ₂ = (3 I) ² Rt 
H ₂ = 9 I ² Rt
In later case, heat generated is 9 times the initial heat generated.
c. If another conductor of 2 R is connected in series 
then total resistance  = R + 2 R = 3 R.
Now heat generated  H = I ² (3 R)t 
                              H = 3I ²Rt
In this case, heat generated is three times.

Question: Explain the following:
(i) Why is tungsten used almost exclusively for filament of electric lamps?
(ii) Why are the conductors of electric heating devices, such as bread-toaster and electric iron, made of an alloy rather than a pure metal?
(iii) Why is series arrangement not used for domestic circuits?
(iv) How does the resistance of a wire vary with its area of cross-section?
(v) Why are copper and aluminium wires usually used for the transmission of electric current?
Answer: (i) It has high resistance and high melting point. So it does not melt when current is passed through it.
(ii) Alloys have more resistivity and higher melting point.
(iii) It increases the resistance and current decreases. Also if one component fails to work, others will also not work
(iv) Resistance is inversely proportional to the area of cross section.
(v) It is because Cu and Al have low resistivity and it allows current to flow.

Question: In a household, 5 tube lights of 40 W each are used for 5 hours and an electric press of 500 W for 4 hour everyday. Calculate the total energy consumed by the tube lights and press in a month of 30 days.
Answer: 
Power of 1 tube = 40W
Power of 5 tubes = 5X40W = 200W

Energy consumed by 5 tubes in 5 hr. per day
                        = 200X5 = 1000Wh

Energy consumed by electric press per day
                                = 500WX4 hr
                                  = 2000Wh

Total energy consumed per day
                             =(1000 + 2000)Wh
                              = 3000Wh = 3 kWh
Total energy consumed in 30 days
                                       = 3X30 kWh
                                      = 90 kWh

Question: In the given circuit, A, B, C and D are four lamps connected with a battery of 60V.Analyse the circuit to answer the following questions.
(i) What kind of combination are the lamps arranged in (series or parallel)?
(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?
(iii) Explain with proper calculations which lamp glows the brightest?
(iv) Find out the total resistance of the circuit. 
Answer: (i) The lamps are in parallel.
(ii) Advantages:
If one lamp is faulty, it will not affect the working of the other lamps.
They will also be using the full potential of the battery as they are connected in parallel.
(iii) The lamp with the highest power will glow the brightest.
P = VI
In this case, all the bulbs have the same voltage. But lamp C has the highest current.
Hence, for lamp C, P = 5 × 60 Watt = 300 W. (the maximum).
(iv) The total current in the circuit = (3 + 4 + 5 + 3) A = 15A
The Voltage = 60V
V = IR and hence R = V/I
R = 60/15 W = 4 W