chapter 11 electricity class 10 science notes

CURRENT ELECTRICITY:-

1. ELECTRIC CURRENT:-

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

Direction of electric current:-

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

Unit of electric current:-

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

I.e.

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

1 mA =10⁻³A, 1 μ A = 10⁻⁶A
An instrument called ammeter measures electric current in a circuit. It is always connected in series in a circuit through which the current is to be measured.
Electric current is a scalar quantity

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

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

Therefore,

Electric charge is

q = 1200 C

Question. A current of 10 A flows through a conductor for two minutes
(i) Calculate the amount of charge passed through any area of cross section of the conductor.
(ii) If the charge of an electron is 1.6 × 10^-19 C, then calculate the total number of electrons flowing. (2013)
Answer: Given that: I = 10 A, t = 2 min = 2 × 60 s = 120 s
(i) Amount of charge Q passed through any area of cross-section is given by

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

$D:\downloads\th.jfif$ Types of current:-

Steady direct current (D.C):-

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

Alternating Current (A.C):-

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

Electric Potential Electric potential is the work done per unit charge in bringing the charge from infinity to that point against electrostatic force. In a conductor, electrons flow only when there is a difference in electric pressure at its ends. This is also called potential difference.

Electric Potential Different

Electric potential difference (pd) between two points in an electric circuit, carrying some current, is the amount of work done to move a unit charge from one point to another.

The S.I. unit of pd is volt (V), where 1 volt

Example: 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

Electric Circuit

A continuous conducting path between the terminals of a source of electricity is called an electric circuit.
A drawing showing the way various electric devices are connected in a circuit is called a circuit diagram.
Some commonly used circuit elements are given below:

Sr. No.	Element	Symbol
1	An electric cell
2	A battery
3	Plug key or switch (open)
4	Plug key or switch (closed)
5	A wire joint
7	Bulb
6	Wires crossing without joining
8	Resistor
9	Variable resistor or Rheostat
10	Ammeter
11	Voltmeter

OHM’s – LAW:- ^m.imp

$C:\Users\vartmaan\Downloads\What-is-the-Ohms-law-1.png$ Ohm’s law states that the potential drop across the ends of a conductor of uniform cross section is directly proportional to the current (I) flowing through it provided constant physical condition such as temperature pressure etc.

Mathematically, V ∝ I

Or V = IR

Or R =

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

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 2 Question. A student plots V-I graphs for three samples of nichrome wire with resistances R₁, R₂ and R₃. Choose from the following the statements that holds true for this graph. (2020)
(a) R₁ = R₂ = R₃
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ > R₁
(d) R₂ > R₁ > R₃
Answer:(d) : The inverse of the slope of I-V graph gives the resistance of the material. Here the slope of –R₂ is highest. Thus, R₂ > R₁ > R₃

(6) Resistance:-^m.imp

It is the property of a conductor to resist the flow of charges through it. Mathematically, it is defined as the ratio of applied potential difference to the current flowing through the conductor.

i,e, R =

Unit of Resistance: –

1 ohm (Ω) = = 1 VA^-1

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

Cause of Resistance:-

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

Symbol of Resistance:-

$C:\Users\vartmaan\Downloads\index.jpg$ $C:\Users\vartmaan\Downloads\images.png$ $C:\Users\vartmaan\Downloads\images.png$

(a) Fixed resistance (b) variable resistance (c) rheostat

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 6 Question. Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph? (2016)

Answer:

Current, I(A)	Voltage, V(V)
0	0
1	2
2	4
3	6
4	8

Resistance is the ratio of values of V and I. Since, the given graph is straight line so; the slope of graph will also give the resistance of the resistor

We can write,

Factors on which the Resistance of a Conductor depends

Resistance of a uniform metallic conductor is:

a) Directly proportional to the length of conductor,

b) Inversely proportional to the area of cross-section,

c) Directly proportional to the temperature and

d) Depend on nature of material.

Resistivity or Factors on which the Resistance of a Conductor depends: –^m.imp

The resistance of a conductor depends upon length l and area A

i.e. (1)

Resistance is inversely proportional to area

i.e. (2)

From equation (1) and (2)

⟹

Here 𝜌 is constant of proportionality called resistivity or specific resistances. Its value depends upon nature of material and temperature.

Definition of Resistivity: –

As R =𝜌

If ,

⇒ R = or = R

Hence resistivity is defined as the resistance of unit cube of substance.

Unit of resistivity: –

As ρ = R = = 1 ohm meter = Ω m.

Resistivity does not change with change in length or area of cross-section but it changes with change in temperature.
Range of resistivity of metals and alloys is to Ωm.
Range of resistivity of insulators is to Ωm.
Resistivity of alloy is generally higher than that of its constituent metals.
Alloys do not oxidize (burn) readily at high temperature, so they are commonly used in electrical heating devices.
Copper and aluminum are used for electrical transmission lines as they have low resistivity.
1 Resistivity depends upon nature of material.
Resistivity for perfect conductors is 0.In perfect conductors there is no resistance at all.
Resistivity for perfect insulators is infinite. There are so many obstacles as a result resistance is more so current cannot flow at all.

Question. A cylindrical conductor of length and uniform area of cross section A has resistance R. The area of cross section of another conductor of same material and same resistance but of length 2 is (2020)

Answer: The resistance of a conductor of length, and area of cross section, A is

Now for the conductor of length 2, area of cross-section and resistivity ρ.

But given, R = R’

⇒ =

or = 2A

Question. Calculate the resistance of a metal wire of length 2m and area of cross section 1.55×10⁶ m², if the resistivity of the metal be 2.8×10^-8 Ωm. (2013)
Answer: given = 2 m
area of cross-section, A = 1.55 × 10^-6 m²
resistivity of the metal, p = 2.8 × 10^-8 Ω m
Since, resistance, = 3.6 × 10^-2Ω or R = 0.036Ω

Question. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10^-8 Ω m. ( 2014)
Answer: given, , = 50 cm = 50 × 10^-2 m. A = 0.01 mm²= 0.01 × 10^-6 m²
ρ = 5 x 10^-8 Ω m.
As, resistance,

= 2.5 Ω

Question. Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms. (2017)
Answer: Given = 1 m, r = 0.01 cm = 1 × 10^-4 m, R = 20Ω

As we know,

∴ 20Ω =

⇒ ρ = 6.28 × 10^-7 Ω m

Question. The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10^-8 ohm meter, find the length of the wire. (2014)
Answer: Here, r = 0.01 cm = 10^-4 m, ρ = 50 × 10^-8 Ω m and R = 10 Ω
As,

⇒= 0.628 m = 62.8 cm

Question. If the radius of a current carrying conductor is halved, how does current through it change? (2014)
Answer: If the radius of conductor is halved, the area of cross-section reduced to (1/4) of its previous value.
Since, , resistance will become four times
From Ohm’s law,
For given V,
So, current will reduce to one-fourth of its previous value.

Question. A wire has a resistance of 16 Ω. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance? (2013)
Answer: When wire is melted, its volume remains same, so,
Here, Therefore,
Resistance, = 16 Ω
Now,
Percentage change in resistance,

Question.A copper wire has diameter 0.5 mm and resistivity 1.6 × 10^-8 Ω m. Calculate the length of this wire to make it resistance 100 Ω. How much does the resistance change if the diameter is doubled without changing its length? (2015)
Answer: Given; ρ = 1.6 × 10^-8 Ω m, diameter of wire, d = 0.5 mm and R = 100 Ω
Radius of wire,

= 0.25 mm = 2.5 × 10^-4 m
Area = 3.14 × (2.5 × 10^-4)²

= 1.9 × 10^-7 m²
As, ⇒= = 1200

If diameter is doubled (), then the area of cross-section of wire will become

Now ,

So the resistance will decrease by four times or new resistance will be

(10) Conductance and conductivity:-

Conductance:- (G) Conductance means something which conducts the current. Greater the resistance lesser the conductance and vice-versa

Unit :- (s) Siemen = ohm^-1(Ω^-1) =mho

Conductivity: – (σ) The reciprocal of resistivity is called Conductivity.

Unit:- ohm metre^-1(Ω^-1) or mho metre^-1siemen metre^-1 (sm^-1)

(11)Combination of Resistance:-^m.imp

(a)Resistance in series

Let us consider three resistors are connected series. Let I is current passing through each resistor then total potential difference applied can be given as (1)

Where (2)

From equation (1) and (2)

If R_s is the equivalent resistance

As V = IR_s

Equivalent Resistance:-

When the three resistors are connected in series, equivalent resistance of the series combination is equal to sum of individual resistance. Equivalent resistance is always greater than largest resistance.

Conclusion:-

 Current through all the resistance (resistor) is same.

 The potential difference across any resistor is proportional to its resistance.

 Current is independent of position of resistor.

(b)Resistors in Parallel:-

Let us consider three resistors are connected in parallel i.e.

Let I is total current passing through resistor.

Than

The potential difference across each resistor is same i.e

If is the equivalent resistance parallel combination

Than

Equivalent Resistance:-

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

Conclusion:-

 Potential difference across the entire resistor is the same.

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

Advantages of Parallel Combination over Series Combination

 In series circuit, when one component fails, the circuit is broken and none of the component works.

 Different appliances have different requirement of current. This cannot be satisfied in series as current remains same.

 The total resistance in a parallel circuit is decreased.

1. The effective resistance of the network between points A and B

(a) 4R (b) 2R



2. Five identical resistances are connected in a network as shown. The resistance measured between A and B is 1 . Each resistance is

(a) 1/4  (b) 4/7 

(c) 7/4  (d) 8/7 

3. What is the effective resistance between A and B

(a) (b)

4. Find the equivalent resistance between P and Q

(a) 10  (b) 5 

5. In the circuit shown,

(a) the pd across 8  is 3.6 V

(b) the pd across 12  is 2.4 V

(d) the current drawn from the battery is 1.5 A

6. In the network shown,

(a) V_AB = +3.0 V (b) V_CB = +6.0 V

Question.If a person has five resistors each of value 1/5 Ω, then the maximum resistance he can obtain by connecting them is
(a) 1 Ω (b) 5 Ω (c) 10 Ω (d) 25 Ω (2020)
Answer:
(a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,

Question. The maximum resistance which can be made using four resistors each of 2 Ω is                                 (2020)
(a) 2 Ω    (b) 4 Ω    (c) 8 Ω  (d) 16 Ω                                                        Answer:
(c) : A group of resistors can produce maximum resistance when they all are connected in series.
∴                                 R_s = 2 Ω + 2 Ω + 2 Ω + 2 Ω = 8 Ω

Question. Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance. (2014)
Answer:
Here, R₁ = 10 Ω, R₂ =15 Ω, R₃ = 5 Ω.
In parallel combination, equivalent resistance, (R_eq) is given by
Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 10

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 12

Question. Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω (2018)
Answer:

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 11

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 13 Question.Three resistors of 3 Ω each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery. (2017)

Answer: equivalent resistance of circuit (R_eq) given by

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

Question. Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases. (2013)
Answer: Let resistance of each resistor be R.
For series combination, So, R_s = R + R = 2R
For parallel combination,
Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 15

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 16 Question . (a) A 6 Ω resistance wire is doubled on itself. Calculate the new resistance of the wire.
(b) Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer.                                               (2020)
Answer: (a) Given, R = 6 Ω            = 6 Ω
Now when the length is doubled, and
∴

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

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 17 Question 35. Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2 V each, a combination of three resistors of 10 Ω, 20 Ω and 30 Ω connected in parallel, a plug key and an ammeter, all connected in series. Use this circuit to find the value of the following:
(a) Current through each resistor
(b) Total current in the circuit
(c) Total effective resistance of the circuit. (2020)
Answer: (a) Given, voltage = 2V + 2V + 2V = 6 V
Current through 10 Ω      I₁₀ = V/R=6/10 = 0.6 A
Current through 20 Ω       I₂₀ = V/R=6/20 = 0.3 A
Current through 30 Ω        I₃₀ = V/R=6/30 = 0.2 A
(b) Total current in the circuit,   1= I₁₀ + I₂₀ + I₃₀  = 0.6 + 0.3 + 0.2 = 1.1 A
(c) Total resistance of the circuit,

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 28 Question. Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, an ammeter and a plug key, all connected in series. Also connect a voltmeter to record the potential difference across the 15 Ω resistors and calculate
(i) the electric current passing through the above circuit and
(ii) potential difference across 5 Ω resistors when the key is closed. (2013)
Answer: V = (2 × 5) V = 10 V
Equivalent resistance,
R_eq = R₁ + R₂ + R₃
= (5 + 10 + 15)Ω = 30 Ω
(i) Current through circuit,

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

V₁ = IR₁ = 0.3 × 5 = 1.5V

Heating Effect of Current

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

Joule’s law of heating effect:

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

Square of the current in the resistor,

H ∝ I² (1)

Resistance of the resistor

H ∝ R (2)

Time for which current flows

H ∝ t (3)

So,

H = I²Rt

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

Filament of electric bulb is made up of tungsten because:

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

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

Practical Applications of the Heating Effects of Electric Current

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

Electric Fuse

It is a safety device that protects our electrical appliances in case of short circuit or overloading.

 Fuse is made up of pure tin or alloy of copper and tin.

 Fuse is always connected in series with live wire.

 Fuse has low melting point.

 Current capacity of fuse is slightly higher than that of the appliance.

Electric Power

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

P = VI = I²R =

The S.I. unit of power is watt (W).
One watt of power is consumed when 1 A of current flows at a potential difference of 1 V.
The commercial unit of electric energy is kilowatt hour (kWh), commonly known as a unit.
1 kWh = 3.6MJ

Units of Electrical Energy:

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

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

Generally, one kwh is called one unit.

Electrical Energy into Mechanical Energy:

Electrical energy can be converted into other forms of energy like heat energy, light energy, motion etc. The best-known examples are:

Fan: The motor in Fan converts electrical energy into mechanical energy
Bulb: Here the electrical energy is converted into light energy.

Example 12.3 (a) how much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?

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

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

(b) Given, V = 220 V, R = 100 Ω.

We have

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

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

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

Question. Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V. (2014)
Answer: Since both the bulbs are connected in parallel and to a 220 V supply, the voltage across each bulb is 220 V.

And

Total current drawn from the supply line,
I = I₁ + I₂ = 0.454 A + 0.273 A = 0.727 A = 0.73 A

Question. How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts. (2016)
Answer: Here, V = 220 V, R = 55 Ω
By Ohm’s law V = IR
∴

Wattage of electric iron =

Power

Question. An electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb. (2015)
Answer: Here,
Power of the bulb

Question. An electric iron has a rating of 750 W; 200 V. Calculate:
(i) the current required. (ii) the resistance of its heating element.
(iii) Energy consumed by the iron in 2 hours. [ 2015]
Answer: Here, P = 750 W, V = 200 V
(i) As so

(ii) By Ohm’s law or
∴
(iii) Energy consumed by the iron in 2 hours
E

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 30 Question 60.An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω. are connected to a 6 V battery as shown in the circuit. Calculate.
(a) The total resistance of the circuit
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp. (2019)
Answer: Resistance of the lamp = 20 Ω
External resistance = 4 Ω
(a) As both the lamp and external resistance are connected in series,

therefore the total resistance, R = 20 + 4 = 24 Ω
(b) Current,

 (ii) Potential difference across conductor



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

Also, resistance of bulb,

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

Also, resistance of the bulb,

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

Question. (a) How two resistors, with resistances R₁ Ω and R₁ Ω respectively are to be connected to a battery of emf V volts so that the electrical power consumed is minimum?
(b) In a house 3 bulbs of 100 watt each lighted for 5 hours daily, 2 fans of 50 watt each used for 10 hours daily and an electric heater of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 3.60 per kWh. (2017)
Answer: (a) Power consumed is minimum when current through the circuit is minimum, so the two resistors are connected in series.
(b) Power of each bulb P₁ = 100 watt
Total power of 3 bulbs, P₁ = 3 × 100 = 300 watt
Energy consumed by bulbs in 1 day

Power of each fan = 50 watt
Total power of 2 fans watt = 100 watt
Energy consumed by fans in 1 day

Energy consumed by heater,

Total energy consumed in one day

Total energy consumed in a month of 31 days

Cost of energy consumed = Rs

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

(b) Energy consumed by electric refrigerator in a day

Question .(i) State one difference between kilowatt and kilowatt hour. Express 1 kWh in joules.
(ii) A bulb is rated 5V; 500 mA. Calculate the rated power and resistance of the bulb when it glows. (2013)
Answer:

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

Resistance of the bulb is