PSEB Solutions for Class 10 Science Chapter 12 Electricity

PSEB Solutions for Class 10 Science Chapter 12 Electricity

PSEB 10th Class Science Chapter 12 Electricity

→ Electric current is the rate of flow of charge Q through the conductor or it is the charge Q flowing per unit time i.e., I = Q/t

→ Charge always flows from a body at a higher potential to a body at a lower potential.

→ Electrostatic potential determines the direction of flow of charge from one body to the other when they are brought in contact.

→ Electrostatic potential at a point is defined as the amount of work done is bringing a unit positive test charge from infinity to that point.

→ Conductors have a large number of free electrons whereas insulators have very few free electrons. The motion of free electrons constitutes the electric current.

→ An electric circuit is a closed path through which electrons flow readily.

→ As per convention, the direction of current is taken as opposite to the direction of the flow of electrons.

→ Electric Current: The rate of flow of charge through a conductor is called electric current.
I = Q/t

→ Ohm: The resistance of a conductor is said to be 1 ohm if a potential difference of 1 volt is maintained between its ends and it allows 1 ampere of current to flow through it.

→ Voltmeter: It is a device used to measure the potential differences.

→ Conductor: Those substances which allow the current to flow through them are called conductors or good conductors.

→ Insulators: Those substances which do not allow the current to flow through them are called insulators.

→ Ammeter: It is an instrument used to measure electric current.

→ Rheostat: It is a device that is used to increase or decrease the current flowing through the electric circuit.

→ One Volt: If 1 coulomb of charge is allowed to pass through a conductor and in doing so 1 joule of work is done then the potential difference across its ends is 1 volt.

→ Ohm’s Law: The ratio of potential difference across the ends of a conductor and the current flowing through it is always constant, provided the physical state of the conductor such as temperature and pressure remains unchanged.
i.e. V ∝ I or V/I = R

→ Electrical Energy: The capacity of doing work by an electric current is called electrical energy.

→ Electric Power: The rate of consumption of electric energy in a conductor is called electric power.

→ Watt: Watt is an S.I. unit of electric energy in which 1 joule of work is done in 1 sec. Kilowatt. It is the power of an agent which can do 1000 joule of work in 1 second.

→ Kilowatt Hour: It is that electric energy which is consumed in an electric circuit in 1 hour.

→ Joule’s Law of Heating Effect. If T is the electric current that flows through a resistor ‘R’ and as a result of which heat produced is ‘H’. Then heat produced is directly proportional to the square of the current and time ‘t’
H = I² Rt

Science Guide for Class 10 PSEB Electricity InText Questions and Answers

Question 1. What is electric circuit?
Answer:
Electric Circuit: The continuous or closed path of electric current from electric source flowing through different components is called electirc circuit.

Components of Electric Circuit: The main components of electric circuit are :

1. Electric source (i.e. battery or cell)
2. Conductor
3. Switch (or key)
4. Any other instrument connected in the circuit.

Question 2. Define uint of electric current.
Answer:
S.I unit of electric current is ‘ampere’ which is denoted by ‘A’.

Ampere: When 1 coulomb of charge flows through a conductor in 1 second, then the electric current flowing through conductor is 1 ampere.
∴ 1A = 1C/ls

Question 3. Calculate the number of electrons that constitute 1 coulomb of charge.
Answer:
We know, the charge present on 1 electron = 1.6 × 10^-19 C
Let ‘n be the number of electrons that constitute 1 coulomb of charge.
∴ n × 1.6 × 10^-19 = 1 C

Question 4. Name a device that helps to maintain a potential difference across a conductor.
Answer:
Electric cell or battery is a device that helps to maintain a potential difference across the ends of a conductor. Chemical action within the cell generates the potential difference across the terminals of the cell.

Question 5. What is meant by saying that the potential difference betwen two points is IV?
Answer:
The potential difference between two points in a current carrying conductor is said to be 1 V if 1 joule of work is done in moving a charge of one coulomb (1 C) from one point to the other.
i.e 1V = 1 J/1C
or 1V = 1 JC^-1

Question 6. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Energy to be given to 1 C of charge to pass through T V potential difference = 1 J.
Energy to be given to 1C of charge to pass through 6 V battery = 6 J or W = VQ
= 6 volt x 1 coulomb
= 6 1J/1C × 1C
= 6 J.

Question 7. On what factors does the resistance of a conductor depend?
                         Or
What is the resistance of a conductor? State the factors on which resistance of a conductor depends?
Answer:
Resistance of a conductor is the opposition offered by it to the flow of electricity through it.

Factors on which resistance of a conductor depends :
Resistance R of a conductor depends upon :

• length of the conductor. It has been experimentally observed that :
  R ∝ l, the length of the wire
• Area of a cross-section of the conductor.
  i.e. R ∝ l/A, where A is the area cross-section of the wire
  Combining above, we have :
  R ∝ l/A
• Nature of material of the conductor.
  or R = ρ l/A
  where p is a constant called specific resistance or electric resistivity of the material of the conductor.

If l = 1, A = 1, ρ = R. .
i. e., specific resistance (or electric resistivity) of a conductor is the resistance of a wire of unit length and a unit area cross-section. It may be defined as :

Specific resistance of a conductor is the resistance of unit cube of the conductor.

Question 8. Will the current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?
Answer:
We know that the resistance of a wire is inversely proportional to the area. Therefore, the thicker the wire, more is the area and lesser the resistance. Thus, current will flow more easily through a thicker wire than through thin wire of the same material.

Question 9. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half the former value. What change will occur in the current through it?
Answer:

• According to Ohm’s law, physical conditions remaining constant, V ∝ I where V is the potential difference across the ends of conductor and I, the current flowing through it.
• When potential difference is halved, the current will also be halved since resistance remain constant.

Question 10. Why are coils of electric toasters and electric irons made of an alloy rather than that of a pure metal?
Answer:
Resistivity (specific resistance) of alloys is much higher than pure metals. Moreover, alloys also do not oxidise readily at very high temperatures. The alloys are, therefore, commonly used in making coils of electric heating devices like electric iron, electric toasters etc.

Question 11. (a) Which among iron and mercury a better conductor?
Answer:
Resistivity of iron is 10 × 10^-8 ohm-m and that of mercury 94 × 10^-8 ohm-m, therefore, iron is a better conductor as compared to mercury.

(b) Which material is best conductor?
Answer:
As silver has low resistivity of 1.60 × 10^-8 ohm-m, so silver is the best conductor.

Question 12. Draw schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor; an 8 Ω resistor and 12 Ω resistor and a plug key, all connected in series.
Answer:
Schematic diagram of the circuit.

Question 13. Redraw the circuit in above question putting an ammeter to measure the current through the resistors and voltmeter to measure the potential difference across 12 Ω resistor. What would be the reading in the ammeter and the voltmeter?
Answer:
Modified circuit is as shown in Figure Since 5 Ω, 8 Ω and 12 Ω are connected in series, therefore, total resistance in series : (R_s) = R₁ + R₂ + R₃
= 5 + 8 +12 = 25 Ω

Now, Total Current through the circuit, I = V/R
= 6/25 = 0.24 A
∴ Reading of ammeter = 0.24 A
P.D. across 12 Ω i.e., reading of voltmeter
V = I × R
= 0.24 × 12.
V = 2.88 Ω

Question 14. Judge the equivalent resistance when the following are connected in parallel (a) 1 Ω and 106 Ω ; (b) 1 Ω and 103 Ω and 106 Ω.
Answer:
(a) Here, R₁ = 1 Ω and R₂ = 106 Ω
When resistances are connected in parallel.

Question 15. 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 the electric iron connected to the same source that takes as much current as all the three appliances and what is the current through it?
Answer:
Combined resistance of 100 Ω, 50 Ω and 500 Ω in parallel i.e., R_p is given by :

Question 16. What are the advantages of connecting electric devices in parallel with the battery instead of connecting them in series?
Answer:
When connected in parallel, all the electric devices are connected to the same potential difference but they draw different current. Moreover, in parallel arrangement if one electric device fails to work, the other devices are not affected. But when connected in series, the potential is divided between the different devices. In case, if one device is switched off or slops working, the other devices also do not work.

Question 17. How can three resistors of resistance 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω ; (b) 1 Ω?
Answer:
(a) When all the three resistors are connected in series, the total resistance will be equal to the sum of all the three resistances.
i.e. R_s = 2Ω + 3Ω + 6Ω
= 11Ω this arrangement is rejected.

(a)
Therefore, when 3 Ω and 6 Ω are in parallel

(b)
R_p’ = 2 Ω. It is in series with 2 Ω.
∴ Total resistance when 3 Ω, 6 Ω in parallel are in series with 2 Ω is = R_p’ + 2 = 2 + 2 = 4Qas given in Figure (b)

Question 18. What is the (a) highest; (b) lowest total resistance that can be secured by combination of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) Total resistance will be maximum when all the four resistances are connected in series.
Let R_s be the total resistance in series combination
∴ R_s =R₁+ R₂ + R_{3 +} R₄
= 4Ω + 8Ω + 12Ω + 24Ω
∴ Highest resistance = 48 Ω

(b) Total resistance will be lowest when all the four given resistance are connected in parallel.
Let Rp be the total resistance in parallel combination,

Question 19. Why does the cord carrying an electric current does not glow while heating element does?
Answer:
Heating effect = 12R where I is the current flowing and R is the resistance.
Current I is the same both in cord and in heating element.

Resistance R of the cord is negligible since it is made of copper (which has very less resistivity) while heating element is made of nichrome whose resistivity is 6,000 times more than copper. Nichrome gets heated up much more than copper. Therefore, heating element glows.

Question 20. Compute the heat generated while transferring 96,000 C of charge in one hour through a potential difference of 50 V.
Answer:
Here, Quantity of charge (q) = 96000 C transferred in 1 hr.
Potential difference = 50 V
Work done to transfer Q coulomb of charge through potential difference W = VQ joule
W = 50 × 96,000
= 48,00,000 J
W = 4.8 × 10⁶ J
∴ Heat produced = Work done in transferring the charge = 4.8 × 10⁶ J

Question 21. An electric iron of resistance 20 Ω takes a current of 5A. Calculate the heat developed in 30 s.
Answer:
Here, Resistance, R = 20 Ω ; I = 5 A ; t = 30 s.
We know, heat produced H = I2 Rt
= 52 × 20 × 30
H = 15,000 J

Question 22. What determines the rate at which the energy is delivered by a current?
Answer:
Rate at which energy is delivered by the current is called power. It is measured in watt or in kilowatt.

Question 23. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Here, current. (I) = 5A
Potential difference (V) = 220 V
Time (t) =2h
= 2 × 60 × 60 s
But Power (P) = V × I .
= 220 × 5
= 1,100 W

Energy consumed in 2 h (= 2 × 60 × 60 s), E = P × t
E = (1,100) (2 × 60 × 60)
∴ E = 7,920,000 J

PSEB 10th Class Science Guide Light Reflection and Refraction Textbook Questions and Answers

Question 1. 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.
Answer:
On cutting wire of resistance R into five equal parts, the resistance of each part is R/5. Let the total resistance be R’ of 5 resistors each of resistance R/5 connected in parallel.

Hence answer (d) 25 is correct.

Question 2. Which of the following terms does not represent electrical power in the circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer:
All three (a), (c) and (d) represent power. Therefore, only (b) IR² does not represent power.

Question 3. An electric bulb is rated 220 V and 100 W. When operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W.
Answer:

Hence (d) 25 W is correct answer.

Question 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of heat produced in series and parallel combination would be
(a) 1 : 2
(b) 2 :1
(c) 1: 4
(d) 4 : 1
Answer:
Since the wires are of the same material and are equal in lengths and diameters, therefore, if R is resistance of each. Let total resistance in series and parallel be Rs and R respectively, when connected in series ;
R_s = R + R
= 2R
When connected to a source V, then

Question 5. How is a voltmeter connected in the circuit to measure potential difference between two points?
Answer:
Voltmeter is always connected in parallel between two points across which the potential difference is required to be measured.

Question 6. A copper wire has diameter 0.5 mm and resistivity 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?
Answer:
Here, Diameter (d) = 0.5 mm
= 0.5 × 10^-3 m
= 5 × 10^-4 m
Resistivity
ρ = 1.6 × 10^-8 Ω m
l = 1

l = 122.76 m
If the diameter of wire is doubled, resistance will become l/4th of the original resistance, since R ∝ 1/A ,
or R ∝ l²d²
Now Resistance R/4=10/4.
= 2.5 Ω
Decrease in resistance will be = 10 – 10/4
= 10 – 2.5
= 7.5 ohm

Question 7. The values of the current I flowing in a given resistor for corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
Graph between I and V is as shown in Figure It is almost a straight line.

Question 8. 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 resistor?
Answer:
Here, I = 2:5 mA = 2.5 × 10^-3 A; V = 12 volts ; R =?
We know from ohm’s law
V = IR

Question 9. 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 will flow through a 12 Ω resistor?
Answer:
Resistors are connected in series
∴ Total Resistance, R8 = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential Difference V = 9 V
Resistance, R = V/I
or I = V/R
= 9/13.4
or I = 0.67 A
When in series, the current passing through each resistor is equal to the current in the circuit.
∴ Current through 12 Ω = 0.67 A

Question 10. How many 176 Ω resistors in parallel are required to carry 5A on a 220 V line?
Answer:
Here R = 176 Ω
V= 220V
I = 5A
Resistance of parallel combination R_p = V/I
R_p = 220/5

If n resistors each of resistance R are connected in parallel, then net resistance R_p is

or n = 4
4 resistors each 176 Ω connected in parallel will result in net resistance of 44 Ω causing a current of 5A to flow through when connected to 220 V.

Question 11. How will you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω ; (ii) 4 Ω?
Answer:
(i) When two resistors each of 6 Ω are connected in parallel give total resistance Rp

∴ 1/6+1/6
= 1+1/6
= 1/3
or R_p = 3 Ω
When this combination is connected in series with third resistor of 6 Ω, it gives a total resistance
(R) = R_p + 6 Ω
= 3Ω + 6Ω
= 9 Ω

(ii) When two resistors each of 6 Ω are connected in series, it gives rise to
R_s = 6 + 6 = 12 Ω

Question 12. 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?
Answer:
Given voltage V = 220 V
Current I = 5A
P = 10 W

∴ n = 110
So, 110 bulbs can be connected in parallel (with each óther across two wires).

Question 13. 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 currents in three cases?
Answer:
1. When two coils A and B each of 24 Ω are used separately
R = 24 Ω and V = 220V
∴ I_sp = 220/24 = 9.17 A

2. When coils A and B are connected in series :
Resultant Resistance R_s = 24 Ω + 24 Ω = 48 Ω
∴Current I_sr = 220/48 = 4.58 A

3. When two coils A and B each of 24 Ω are connected in parallel

Question 14. Compare the power used in 2 Ω resistor in each of the following circuits :
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors and
Answer:
Here R₁ = 1 Ω ; R₂ = 2 Ω
when connected in series the resultant resistance, R = R₁ + R₂
= 1 Ω + 2 Ω.
= 3 Ω
∴ Current through each resistor and also total current
I = V/R_s
= 6/3
= 2A

Power used in 2 Ω resistor P₁ = I² × R
= (2)² × (2)
= 4 × 2
= 8 W
i.e Power P₁ = 8W

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
Let 2 Ω and 12 Ω in parallel combination give resultant resistance R

Current divides among its parallel resistances in the inverse ratio of resistances.

It is seen that P₁ = P₂ = 8W .
The Power consumed by 2 Ω resistor in each case is 8 W.

Question 15. Two lamps, one rated loo W at 220 V and other 60 w at 220 V, are connected in parallel to electric main supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Given Power, P = V²/R :
or R = V²/P

Question 16. Which uses more energy, a 250 W TV set for 1 hour or a 1,200 W toaster for 10 minutes?
Answer:
(i) Energy consumed when TV set of 250W is used for 1 hr. .
E₁ = P₁ × t²
= 250 W × l h
∴ E₁ = 250 Wh

(ii) Energy consumed when 1200 W toaster is used for 10 minutes.
= 1,200 W × 10/60 hr
E₂ = 200 Wh.
From (i) and (ii) E₁ > E₂
Therefore, 250 W TV set consumes more energy than a toaster of 1200 W.

Question 17. An electric heater of resistance 8 Q draws 15 A from service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
Given R = 8 Ω, I = 15A and Time t = 2 hrs.
Rate at which heat is developed in the heater, its electric power,
P = I² × R
= 152 × 8
= 225 × 8
i.e P = 1,800 W or Js^-1
Thus, 1,800 joule of heat is developed per second

Question 18. Explain the following :
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
Answer:
Melting point of tungsten is very high and also its resistance is high. Hence tungsten can be heated to very high temperature. Due to its high resistance, large quantity of heat is produced which makes it glow.

(b) Why are the conductors of electric heating devices, such as bread toasters and electric irons, m^e of an alloy rather than pure metal?
Answer:
Resistivity of alloys is very large as compared to pure metals. Moreover, alloys donot/Oxidise (burn out) easily as compared to pure metals. Therefore, conductors of electric heating devices are made of alloys.

(c) Why is series arrangement not used in domestic circuits?
Answer:
When electric appliances are connected in series,’ the equivalent resistance will be very large as
R_eq = R₁ + R₂ + R₃ + R₄ + ………. Due to high equivalent resistance huge quantity of heat is produced in the domestic circuit which may result in fire. This is why series arrangement is not used in domestic circuits.

(d) How does the resistance of a wire vary with its area of cross-section?
Answer:
Resistance of a conductor is inversely proportional to its area cross-section.

(e) Why are copper and aluminium usually employed for electricity transmission?
Answer:
Copper and aluminium have low resistivity and as such there will be small loss of energy when a certain current flows through them. Therefore, copper and aluminium wires are used for electricity transmission.

PSEB 10th Class Science Important Questions Chapter 12 Electricity

Multiple Choice Questions:

Question 1. V ∝ I law was given by:
(A) Faraday
(B) Watt
(C) Ohm
(D) Coulomb.
Answer:
(C) Ohm

Question 2. The unit of Potential is:
(A) Ampere
(B) Volt
(C) Ohm
(D) Watt.
Answer:
(B) Volt

Question 3. The unit of electric energy is:
(A) Ampere
(B) Volt
(C) Ohm
(D) Watt
Answer:
(D) Watt

Question 4. Resistance of a conductor depends on:
(A) its length
(B) its area of cross section
(C) nature of its material
(D) All of these.
Answer:
(D) All of these.

Question 5. What maximum resistance can be obtained by combining three resistances each of ___________
(A) 1/3 Ω
(B) 1Ω
(C) 1/9 Ω
(D) 3Ω.
Answer:
(B) 1Ω

Question 6. By which unit electric current is represented?
(A) Coulomb
(B) Ampere
(C) Watt
(D) Kilowatt.
Answer:
(B) Ampere

Question 7. Electric current in circuits is measured by:
(A) Ammeter
(B) Voltmeter
(C) Galvanometer
(D) Electric meter.
Answer:
(A) Ammeter

Question 8. How is Ammeter always connected in circuits? :
(A) in series
(B) in parallel
(C) both in series and parallel
(D) None of these.
Answer:
(A) in series

Question 9. How much work is done to carry 2 C of charge between two points?
(A) 2 J
(B) 6 J
(C) 24 J
(D) J.
Answer:
(C) 24 J

Very Short Answer Type Questions

Question 1. Define energy.
Answer:
Energy. It is the capacity to do work.

Question 2. Define electric energy.
Answer:
Electric Energy. Work done is the electric energy used to produce heat energy in an electric circuit.

Question 3. Define electric current and state its unit.
Answer:
Current. It is rate of flow of electric charge. Its unit is ampere.

Question 4. Define a volt, whose unit is this?
Or
What is P.D.? Give SI unit.
Answer:
P.D. (Volt) is work done in moving 1 coulomb of + ve charge from one point to other. It is the unit of potential difference. SI unit of P.D. is JC^-1 or volt.

Question 5. Show the switch signs in circuit in (i) open (ii) closed circuit.
Answer:
Circuit Sign

Question 6. Is electric potential a scalar or a vector quantity?
Answer:
It is a scalar quantity.

Question 7. What is practical unit of power and electric energy?
Answer:
Practical unit of power is watt and that of electric energy is kWh (kilo watt hour).

Question 8. Which one is having more resistance, 100 W bulb or a 50 W bulb?
Answer:
Resistance of 50 W bulb is twice that of 100 W bulb.

Question 9. What constitutes the current?
Answer:
Flow of free electrons constitute the current.

Question 10. What is SI unit of resistivity?
Answer:
It is ohm-m.

Question 11. What is conductor of electricity? Give two examples.
Answer:
Conductor of electricity. A substance that allows the electric current to pass through it is called conductor of electricity.

Examples :

• Copper
• Silver
• Human body.

Short Answer Type Questions

Question 1. What is the potential difference between two points in the electric field? Name and define its SI unit.
Answer:
Let a charge Q be moved from one point to another point, in the electric field and W be the work done, then the potential difference V between two points is given by :
V = W/Q
or W = VQ
If Q = + 1C, then V = W

Definition: The potential difference between two points in the electric field is defined as the amount of work done in moving a unit positive test charge from one point to the other against electrostatic force due to electric field.

SI units of P.D. is volt: Potential difference between two points is said to be 1 volt if 1 J of the work is done in moving a charge of 1 C from one point to the other.
Hence 1 volt = 1 joule / 1 coulomb 
1 V =1 J/C
= 1 JC-1

Question 2. What is an electric circuit?
Answer:
Electric circuit. It is a closed path through which the electrons flow readily. When a conductor is connected to a battery, the electrons move from negative terminal of the battery to the positive terminal. However, conventionally, the current is considered to flow from positive to negative terminal of battery.

Question 3. Distinguish between good conductors, resistors and insulators.
Answer:

• Good Conductors: Those substances through which the current can flow freely.
• Insulators: Those substances which do not allow the current to flow through them.
• Resistors: These are objects which oppose the flow of current through them.

Question 4. What is the contribution of electricity in our daily life?
Answer:
Contribution of Electricity in our Life. Electricity finds great contribution in our life. It gives many facilities like its use in electric bulb and tubelight removes darkness at night, in summers with its use desert coolers and air conditioners keep our houses cool while in winters with its use in heaters we keep our houses warm. Besides this electricity is used in televisions, radios and cinemas for our entertainment. Electricity is also used to work the various machines the fields of agriculture, transport and industry.

Question 5. What do you understand by static electricity?
Answer:
Static Electricity: When two bodies are rubbed against each other then they acquire a property to attract lighter bodies towards them that is they get static charge. Electricity produced by rubbing the two bodies with each other is called frictional electricity or static electricity. And the study of static charges is called electrostatics.

Example: When a plastic pen is rubbed with dry hair then the pen attracts small bits of paper towards itself. This is due to the electricity produced by rubbing.

Question 6. What are positive and negative charges? How are these produced?
Answer:
Positive charge: The charge produced in glass rod when it is rubbed with silk cloth is called positive charge.
Negative charge: The charge produced on ebonite rod when it is rubbed with cat skin is called negative charge.

Question 7. Name and define unit for electric current?
Answer:
The unit of electric current is Ampere.
Ampere: If 1 coulomb of charge is allowed to flow through a conductor in 1 second, then the current flowing through the conductor is said to be 1 ampere.

The smaller unit of current is milli 1 ampere
1 milliampere = 1/1000 ampere
= 10^-3 ampere

Long Answer Type Questions

Question 1. (a) What is meant by Joule’s heating effect due to flow of current through a conductor?
Answer:
The conductors offer some resistance to the flow of current. If I is the current in ampere flowing for t second, then quantity of charge Q = I × t.

The work done in carrying a charge Q ( It). coulomb against a potential difference of
1 volt. W = V × Q
or W = V × It
= VIt joule

The whole of this energy is converted into heat.
∴ Heat produced, H = W = VIt joule ……….. (i)

(b) Define electric power and unit of electric power.
Answer:
Electric Power: The rate of doing electric work is called electric power. Suppose ‘V’ is the potential difference between the ends of a conductor and T is the current flowing through it,

Work done by flow of current T for ‘t’ second, W = VIt

Unit of Electric Power. We know P = V x I. If potential difference V is measured in volt and current T in ampere than power will be in watt.
1 watt = 1 volt x 1 ampere

The bigger unit of power is Kilowatt (kw)
1 kilowatt = 1000 watt.

(c) What do you mean by electric energy? Give the definition of its unit.
Answer:
Electric Energy: Total work done by electric current in a fixed time is called electric energy.

Suppose I ampere of current flows through a conductor for t seconds when the potential difference across the ends of a conductor is V, then total work done or electric energy consumed,
E = W = V × I × t

The unit of electric energy is joule or watt-second but this is a small unit.
The bigger unit of electric energy is watt hour.

Watt hour. The electric energy consumed in a circuit is said to be watt hour if 1 ampere of current flows for 1 hour and potential differenc across the ends is 1 volt.
1 watt-hour = 1 watt × 1 hr.
= 1 volt × l ampere × 1 hour 1 kilowatt hour (kwh) = 1000 watt hours
The bigger (commercial) unit of electric energy is kilowatt hour (kwh).

Question 2. What is meant by resistance of a conductor? On what factors does the resistance of conductor depends?
Answer:
Resistance of Conductor. It is defined as the ratio of potential difference across its ends to the current flowing through the conductor is called resistance of the conductor. It is denoted by ‘R’.

If V is the potential difference between the ends of the conductor and T is the current flowing through the conductor, then
Resistance (R) =  Potential Difference (V) / Current (I)

Unit of Resistance. S.I. unit of resistance is ohm.

Factors on which resistance of conductor depends.

• Electric source (i.e. battery or cell)
• Conductor
• Switch (or key)
• Any other instrument connected in the circuit.

Question 3. Find experimentally the various factors on which resistance of conductor depends.
Answer:
On what factors the resistance of a conductor depends is shown by the following experiment :

Experiment: Arrange an electric circuit consisting of a battery, ammeter, connecting wire and conductor with the help of a switch (key). Press the switch and allow the current to flow through the circuit. Note the value of current from the ammeter. Now in place of this wire, connect another wire of place of this wire, connect another wire of same length and thickness and note the reading of ammeter. You will find that the value of current changes. This experiment shows that resistance of conductor depends upon the nature of material of wire i.e. at the same temperature the resistance of different conductors of same length and thickness is different.

Now take a wire of the same material as that of the first conductor and same diameter but of double length. Connect this wire in the same circuit and allow the current to flow through it. You will see this value is half of the first value. This shows that resistance is proportional to the length. If R is the resistance of the conductor and l the length then,
R ∝ l …(i)
Now take two wires of the same material but of different areas of cross section. At first connect a wire of small cross-section in the circuit and then replace it with a wire of large area of cross-section. Note the value of current I in two cases. You will see that more current flows in the second wire than in the first wire. It is clear that resistance of second wire is less than the first wire.
i.e. R ∝ 1/A ……(ii)

Combining (i) and (ii), we get
R ∝ l/A …..(iii)
or R = ρ. l/A …(iv)
where ρ is a constant and is called resistivity of the conductor. Its value depends upon the nature of material of conductor (wire).

Question 4. What is Ohm’s law? How can it be verified?
                              Or
Write ohm’s law. Draw a circuit diagram to prove it experimentally in the laboratory,
Answer:
Ohm’s law. It states that the current passing through a conductor is directly proportional to the potential difference across its ends provided the temperature and other physical conditions remains unchanged.
I ∝ Y
i.e. V ∝ I
or V = RI
Where R is a constant of proportionalty called resistance.

Resistance. It is the property of a conductor to oppose the flow of current. Resistance depends upon the nature of the conductor, its temperature and its dimensions (length, area).
R = V/I
or I = V/R
Experimental Verification of Ohm’s Law: Connect an ammeter, battery, key and rheostat as shown in Fig. 12.7(a). Put the plug in the key K. Read the values of potential difference across resistor R with the help of voltmeter V and the current flowing through the resistor with the help of ammeter.

Note the readings. Vary the current in the circuit by sliding contact of rheostat and go on noting reading in voltmeter and ammeter. Take the ratio potential difference (V) and corresponding value of current (I) each time. The ratio would be same, which proves Ohm’s law. Now plot a graph between V and I on graph paper. It will be straight line graph as shown in Figure (b) which also verifies that V ∝ I.

Question 5. What is the need of combining different resistors? What is the resultant resistance when a number of resistances are connected in series?
                                            Or
State Ohm’s law. Find the equivalent of resistances of the individual resistances connected in series.
Answer:
Ohm’s law: It states that the current passing through a conductor is directly proportional to the potential difference across its ends provided the temperature and other physical conditions remains unchanged.
I ∝ Y
i.e. V ∝ I
or V = RI
Where R is a constant of proportionalty called resistance.

Combination of Resistors: Resistors of all values of resistances are not available. Hence resistors are connected in a number of ways to increase or decrease the combined resistance. There are two distinct ways in which resistors can be connected. They are (i) resistors in series and (ii) resistors in parallel.

Resistors Connected In Series:

Resistors are said to be connected in series, if they are joined end to end so that the same current flows through each one of them in succession. Let the resistors R1( R2 and R3, be joined in series and let the current passing through them be I [Figure].

Let V₁, V₂, and V₃ be the potential difference between the ends of the first, second and third resistor respectively.
By Ohm’s law,
V₁ = IR₁; V₂ = IR₂ and V₃ = IR₃

If V is the total potential difference between the ends A and D and R is the effective resistance of the combination of all the resistors, then
V = IRS
But V = V₁ + V₂ + V₃.
or IR_s = IR₁ + IR₂ + IR₃
IR_s = I(R₁ + R₂ + R₃)
or R_s = R₁ + R₂ + R₃

The above result holds good for any number of resistors joined in series.
Thus when resistors are joined in series, the total resistance is equal to the sum of individual resistances.

Question 6. With the help of a diagram derive the formula for the equivalent resistance of three resistances connected in parallels.
                             Or
With the help of a labelled circuit diagram derive a formula to find combined resistance (R) when two or more resistances (R₁, R₂, R₃) are connected in parallel taking symbols potential difference (V) and current (I).
Answer:
A circuit in which two or more resistors are connected across two common points so as to provide separate paths is called parallel circuit.

In this case, the same potential difference will be maintained between the two ends of every resistor and the current will divide itself in various branches.

Let the resistors R₁; R₂ and R₃ be joined in parallel between the points A and B.
Let the current I reaching A divide itself into three parts I₁; I₂ and I₃ along R₁, R₂ and R₃ respectively. Let V be the potential difference between the points A and B.

Question 7. What is meant by electric power? Give its units. Also give its SI unit.
                     Or
What is power? Give commercial unit of power.
                    Or
What is electric energy? What is its SI unit?
Answer:
Power is the rate of doing work. The electric power of an appliance is its rate of consumption of electric energy.
The power of an electric appliance is 1 W (watt) if it consumes 1 J (joule) of energy in 1 second.

If I ampere of the current flows for t second through a coil whose ends are maintained at a potential difference of V volt, then the energy consumed or the work done.
W = VIt joule

Electric energy is the total amount of work done by the current in a given time or electric energy is the total amount of energy consumed in an electric circuit in a given time.
W (joule) = P(watt) × t (second)
or W = P(J/s) × t (s)
= P × t joule
But P = VI
∴ W = VIt joule

Practical Unit of Electric Energy: Practical unit of electric energy called kWh (kilowatt-hour) is usually used. This unit is also called B.O.T. (Board of Trade Unit).

The unit kWh is equal to the work done or energy consumed when a power of IkW is consumed for 1 hour.

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