3.Electricity-Full
Q.1 Give the concept of electric charge briefly.
Ans. Concept Of Electric Charge:– Electric charge or charge, like mass is a fundamental property of matter. We know that, when two substances are rubbed with each other they attain a property of attracting small bits of paper or straw etc. Acquiring this attractive power on rubbing by the two bodies are said to have become electrified or acquire electric charges.
e.g. If a comb is brought near some tiny pieces of paper, it does not attracts them, but on rubbing first with hair, it starts attracting them towards itself. This implies that, initially, the comb is electrically neutral so it does not attract tiny pieces of paper but on rubbing, it gets electric charges and thus attracts them.
The S.I. Unit of electric charge is Coulomb and is denoted by ‘C’.
1C = 6.25×1018 electrons
Q.2 What are the two types of electric charges?
Ans. Types of Electric Charges:– It has been found by experiments that there are two types of electric charges-positive and negative charges. The charge carried by a proton is positive while that carried by an electron is negative. A body gets positively charged if it loses electrons and negatively charged if it gains electrons. E.g. When a rod is rubbed with silk, it loses electrons and thus becomes positively charged while silk on other hand gains an equal number of electrons and thus becomes negatively charged. A proton possess a positive charge of 1.6×10-19 C while as an electron possess a negative charge of 1.6×10-19 C
Q.3 List some properties of Electric charges.
Ans. Properties of Electric Charges:- Some properties of electric charge are as:
1. Like charges repels each other while unlike charges attracts each other.
2. Electric charge is conserved i.e. it can neither be created nor destroyed.
3. The force (F) between two charges varies directly as the product of two charges (q1 & q2) and inversely proportional as the square of the distance (r) between them.
i.e.
q1 x q2
F α ───────
r2
q1 x q2
ðF = k ──────
r2
Where k = constant of proportionality.
4.Electric charge is quantized. By quantization, we mean that the charge (Q) on any body is integral multiple of electronic charge i.e.
Q = ne
Where n = ±1, ±2, ±3,……….
e = 1.6×10-19 C
5. Electric charge is additive i.e. total charge is the algebraic sum of the individual charges.
Q.4 Define electric potential and potential difference?
Ans. Electric Potential:- The amount of work done when a unit positive charge is brought from infinity to a given point in an electric field is called electric potential or potential.
The electric potential energy per unit charge is also electric potential.
Electric potential energy
:. Electric Potential = ──────────────────────
Charge
Its unit is Volt and is denoted by ‘V’
Potential difference:- The potential difference between two points in an electric field is the work done in taking a unit positive charge from one point to another against the electric field.
The difference in electric potential between two points is known as potential difference.
:. Potential difference = Work doneQuantity of charge moved
ð V = WQ
Where, W = work done
And Q = Quantity 0f charge moved
The S.I. unit of potential difference is volt and is denoted by ‘V’. It is measured by an instrument called voltmeter.
Q.5 Define coulomb and volt.
Ans. Coulomb:- One coulomb is that quantity of electric which exerts a force of 9×1010 Newton on an equal charge placed at a distance of 1 meter from it.
Volt:- One volt is the potential difference between two points in an electric field when one joule of work must be done to move one coulomb of charge between the two points.
i.e. 1 Volt = 1
joule1
coulomb
ð1 V = 1 J/C
Q.6 Define electric current. What is its S.I. unit?
Ans. Electric Current:- The ordered motion of electric charges (called electrons) in a conductor is known as electric current. The rate of flow of electric charge through any cross-section of a conductor is called the strength of a current or magnitude of current.
If ‘Q’ coulombs be the charge flowing across any cross-section of a conductor in time ‘t’ seconds, then the current (I) is given by
I=Qt
The S.I. unit of electric current is Ampere and is denoted by ‘A’.
It is measured by an instrument called ammeter.
Q.7 Define ampere.
Ans. Ampere:- We know that
I=Qt
Where Q = coulombs of electric charge.
And t = time.
If Q = 1C
& t = 1 sec.
Then, I = 1C/1sec.
ðI = 1 C/sec.
ðI = 1 A
Hence, current flowing through a conductor is said to be one ampere if one coulomb of charge flows through it in one second.
Q.7 Explain briefly how the electric charges flows in a wire?
Ans. We know that a metallic conductor has a plenty of free electrons in it and these free electrons move at random in all directions within the conductor. But, when the ends of the conducting wire are connected to a source (cell or battery) these free electrons starts moving from positive end to the negative end. This flow of electrons constitutes electric current in the wire. The average speed with which the electrons move (or drift) in a wire having potential difference across its ends, is called the drift speed. See fig. below
Q.8 What is the conventional direction of the flow of electric current?
Ans. Direction of Electric Current:- By convention, the direction of electric current is taken as the direction of flow of positive charges from a point at a higher potential to a point at a lower potential. However, in metallic conductors negative electrons move not the positive charges. The flow of electrons is in the direction opposite to that in which the conventional current flows.
Q.9 State Ohm’s law. How will you verify it experimentally?
Ans. Ohm’s Law:– Ohm’s law gives a relationship between electric current and potential difference. According to this law, “ At constant physical conditions like temperature etc., the current flowing through a conductor is directly proportional to the potential difference across its ends.”
If ‘I’ is the current flowing through a conductor and ‘V’ is the potential difference across its ends, then according to Ohm’s law:
I α V
ðV α I
ðV = I x R
Where ‘R’ is a constant and is called the resistance of the conductor whose value depends on size, nature of the material and temperature of the conductor.
Experimental verification of Ohm’s Law:- To verify Ohm’s law, let us take a conductor XY whose ends are connected to a voltmeter (V). Also connect an ammeter (A), battery (B), Key (K) and rheostat (Rh) as shown in the fig.
Now, press the key (k), a current starts flowing in the whole circuit including the conductor ‘R’. Note the current ‘I’ in the circuit by the ammeter and the potential difference (v) across the ends of a conductor by voltmeter. By changing the current with the help of rheostat, take several readings of current as well as potential difference. Now on calculating value of V/ I for all the readings, we observe that each time the ratio of potential difference and current is constant. This verifies Ohm’s law.
If a graph between potential difference readings and corresponding current readings is drawn, we will get a straight line graph which shows that the current is proportional to the potential difference. This also verifies Ohm’s law.
Q.9 What do you mean by resistance of a conductor? What is its S.I. unit?
Ans. Resistance:- The electric current is a flow of electrons through a conductor. When the electrons move from one part of the conductor to other part, they collide with each other and with other atoms & ions within the conductor. Due to these collisions, there is some obstruction or opposition to the flow of current through the conductor. The property of a conductor due to which it opposes the flow of current through it, is called resistance of the conductor.
The ratio of the potential difference across its ends to the strength of the current flowing through it, is called the resistance.
i.e. R = VI
The S.I. unit of resistance is Ohm and is denoted by ‘Ω’
Q.10 Define Ohm.
Ans. We know that
R = V/I
If V = 1 volt
& I = 1 ampere,
Then, R = 1V /1A
ðR = 1 V/A
ðR = 1 Ohm
ðR = 1 Ω
Hence, the resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is applied across its ends.
Q.11 On what factors does the electrical resistance of a conductor depends?
Ans. The electrical resistance of a conductor depends upon the following factors:
1. Length of a conductor:- It has been found by experiments that the resistance of a wire increases on increasing its length and decreases by decreasing its length. In other words, the resistance of a conductor is directly proportional to its length.
i.e. R α l
Where l is the length of conductor.
From the above relation, it is clear that if the length of the wire is doubled, its resistance gets doubled and if it is halved, its resistance also gets halved.
2. Area of Cross-Section:- It has been found by experiments that the resistance of a wire is inversely proportional to its area of cross-section.
I.e. R α 1A
Where A = area of cross-section
From the above relation, it is clear that if the area of cross-section of the wire is doubled, its resistance gets halved and vice-versa.
3. Temperature of the conductor:- It has been found by experiments that the resistance of all pure metals increases on increasing the temperature and decreases on lowering the temperature. But, the resistance of alloys like nichrome, constantan, etc is almost unaffected by temperature.
4. Nature of the material:– The electrical resistance of a conductor depends on the nature of the material of which it is made. Some materials have low resistances where as others have high resistances. e.g. In copper metal and nichrome wire with equal lengths and diameters, nichrome alloy has more (60 times) resistance than the copper metal.
Q.12 What do you mean by resistivity? What is its S.I. unit?
Ans. We know that
R α I ①
Where ‘l’ is the length of the conductor.
Also,
R α 1A ②
Where, A = area of cross-section.
From ① and ②, we get
R = lA
ð R = plA
Where p (rho) is a constant and is known as resistivity of the conductor.
Resistivity is also known as specific resistance.
If A = 1m2, l=1m.
Then, R = px11
ðR = p
Hence resistivity is the resistance of that substance which is 1m in length and 1m2 in cross-section (thickness).
The S.I. unit of resistivity is Ohm-meter denoted by Ω m.
Q.13 Derive the formula for the equivalent resistances of three resistors connected in series.
Ans. Resistances connected in series:- When two or more resistances are connected end to end consecutively, they are said to be connected in series. In this case, the total resistance increases.
Let R1, R2, and R3 be three resistors connected in series and V1, V2, and V3 be the potential differences across these resistances. If ‘V’ be the potential difference due to battery, then
V = V1 + V2 + V3 ①
Let the current ‘I’ flows through the conductor whose combined resistance is ‘R’,
Then by the Ohm’s law, we have
V = I R
Again applying Ohm’s law to each resistance separately, we have
V1 = I R1
V2 = I R2
V3 = I R3
Substituting the values of V, V1, V2 and V3 in ①, we get
I R = I R1 + I R2 + I R3
ðI R = I x (R1 + R2 + R3)
ðR = R1 + R2 + R3
Hence, equivalent resistance R is equal to the sum of individual resistances R1, R2 and R3 connected in series.
Q.13 Derive the formula for the equivalent resistances of three resistors connected in parallel.
Ans. Resistances connected in parallel:- When two are more resistances are connected between the same two points, they are said to be connected in parallel. In this case, the total resistance decreases.
Let R1, R2 and R3 be the three resistors connected in parallel, V is the potential difference across the ends of each conductor and I1, I2 and I3 are currents flowing through these resistances respectively. Let R be the equivalent resistance, then
I = I1 + I2 + I3 ①
According to Ohm’s law
V = I R
ðI = VR
Similarly, applying Ohm’s law to each resistance separately, we get
I1= VR1
I2 = VR2
I3 = VR3
Substituting the values of I, I1, I2 and I3 in① , we get
VR=VR1+VR2+VR3
ðV1R=V(1R1+1R2+1R3)
ð1R=1R1+1R2+1R3
Thus, the reciprocal of the equivalent resistances is equal to the sum of the reciprocals of the individual resistances when connected in parallel.
Q.14 What are the disadvantages of series circuits for domestic wiring?
Ans. In domestic circuits, series arrangement is not used because of the following disadvantages:
1. If one of the appliance stops working due to some defect, then all other appliances also stops working.
2. All the appliances have only one switch due to which they con not be turned on or off separately.
3. All the appliances do not get the same voltage (220V) as that of power supply line because the voltage is shared by all the appliances.
4. The overall resistance of the circuit increases due to which the current from the power is low.
Q.15 What are the advantages of parallel circuits in domestic wiring?
Ans. In domestic wiring, parallel combination is used because of the following advantages:
1. In parallel circuits, if one electrical appliance stops working due to some defect, then all other appliances keep working normally.
2. In parallel circuits, each electrical appliance has its own switch due to which it can be turned on or off independently, without affecting other appliances.
3. In parallel circuits, each electrical appliance gets the same voltage (220V ) as that of the power supply.
4. In parallel combination, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.
Q.16 Define electric power? What is its S.I unit?
Ans. Electric Power:- The rate at which electric work is done by an electric current is called electric power.
i.e. Electric Power = Electric
Work doneTime Taken
or P=Wt
Electric power is also defined as the rate at which electrical energy is consumed.
i.e. Power=Energy
consumedTime
taen
The S.I. unit of electric power is watt.
Formulae for electric power:
We know that
Power=Work
doneTime
Taken
ð P=Wt ①
Since the work done 
by current ‘I’ when it flows for time ‘t’ under a potential difference ‘v’ is given by
W = V x I x t joules
Substituting this value in ①, we get
P=V x I x tt
ðP=V x I watts
Q.17 Define electric energy. What is its commercial unit?
Ans. Electric Energy:– The total work done by a current in an electric circuit is called electric energy.
We know that
Electric Power=Work
doneTime
taken
Since, work done by
current=Electric energy consumed
:. Electric Power = electrical
energy consumedTime taken
Or Electrical Energy = Power x Time
ðE = P x t
Hence, electrical energy consumed by an electrical appliance is the product of its power rating and time for which it is used.
Commercial Unit of Energy:- The S.I. unit of electrical energy is Joule. Actually, joule represents a very small quantity of energy, so for commercial purposes a bigger unit called kilowatt-hour is used.
One kilowatt- hour (kWh) is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for one hour.
:. 1 kilowatt-hour=1kilowatt
x 1 hour
ð1kwh= 1000watts x 1hour
ð1kwh = 1000 watts x 60 x 60 sec.
ð1kwh = 1000 J/sec. x 3600 sec.
ð1kwh = 3600000 Joules
ð1kwh = 3.6 x 106 J
Q.18 What do you mean by heating effect of current?
Ans. Heating Effect of Current:- When an electric current is passed through a high resistance wire, it becomes very hot and produces heat. This is called the heating effect of current. The heating effect of current is obtained by the transformation of electrical energy into heat energy.
Consider a resistance wire with resistance ‘R’ through which a current ‘I’ flows in time ‘t’ and let an electric charge ‘Q’ moves against a potential difference ‘V’, then
Work done = Q x V ①
We know that
Current, I = Q/t
ðQ = I x t
Also, from Ohm’s law, we have
V = I x t
Substituting the values of Q and V in ①, we get
W = I x t x I x R
ð W = I2 x R x t
But this work done = Heat Produced(H)
:. H = I2 x R x t Joules
This is known as Joules law of heating. It is clear from the joules law of heating that heat produced is directly proportional to:
a. time (t).
b. Square of current (I2).
c. resistance of wire (R).
Q.19 List some important applications of heating effect of current.
Ans. Some important applications of heating effect of current are as:
1. The heating effect of current is used in electrical appliances like electric iron, electric kettle, electric oven, room heaters etc.
2. The heating effect of current is used in electric bulbs for producing light.
3. The heating effect of current is used in electric fuse for protecting household wiring and other electrical appliances.
Numerical Problems
Problem 1: If four resistances, each of value 1 ohm, are connected in series, what will be the resultant resistance?
Sol.
Here, R1 = R2 =R3 = R4 = 1 ohm.
Now, for resistances connected in series, we have
R = R1 + R2 + R3 + R4
ðR = 1 + 1 + 1 + 1
ðR = 4 Ohm
ðR = 4Ω
Hence, the resultant resistance is 4Ω
Problem 2: Calculate the equivalent resistance when two resistances of 3 ohm and 6 ohm are connected in parallel.
Sol.
Here, R1 = 3 ohms and R2 = 6 ohms.
Now for resistances connected in parallel, we have
1R=1R1+1R2
ð1R=13+16
ð1R=2+16
ð1R=36
ð1R=12
ðR = 2Ω
Hence, equivalent resistance is 2 Ω.
Problem 3: What will be current drawn by an electric bulb of 40W when it is connected to a source of 220V ?
Sol.
Here, Power (P) = 40watts,
Voltage (V) = 220Volt
Now, we know that
P = V x I
ð40 = 220 x I
ðI=40220
ðI=0.18 ampere.
Hence, current drawn by electric bulb = 0.18 ampere.
Problem 4: A radio set of 60watts runs for 50 hours. How much electrical energy is consumed.
Sol.
Here, Power (P) = 60watts = 601000 kilowatts = 0.06 kw.
And, Time (t) = 50 hrs.
Now, we know that
E = P x t
ðE = 0.06 x 50 = 3 kilowatt-hours
Hence, Electrical energy consumed by a radio set = 3 KWh.
Problem 5: A bulb is rated at 200V – 100W. Five such bulbs burn for 4 hours.
(a)What is the electrical energy consumed?
(b) Calculate the cost if the rate is Rs 4.60 per unit?
Sol.
Here, Power (P) = 100watts = 1001000kilowatts=0.1kw
Voltage (V) = 200Volts
& Time (t) = 4 hrs.
(a).We know that
E = P x t
ðE = 0.1 x 4 = 0.4 KWh
ðEnergy consumed by 1 bulb = 0.4 KWh
.: Energy consumed by 5 bulbs = 0.4 x 5 = 2 KWh.
Hence, total energy consumed = 2KWh = 2 units.
(b).
Since cost of 1 unit of electricity = Rs. 4.60
.: Cost of 2 units of electricity = 4.60 x 2 = Rs 9.20
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