ELECTRICITY PART 3

ELECTRICITY PART 3

PREVIOUS TOPICS:

ELECTRICITY PART 1
ELECTRICITY PART 2

RESISTANCE (R) :

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

Resistance of a conductor is the opposition offered to the follow of electric charge in a conductor.

OR

It is the ratio of the potential difference across the ends of the conductor to the current flowing through it.

R=V/I

S I Unit of resistance is ohm (Ω).

1 Ω = 1 V/1 A = 1 V A^-1

Definition of 1 ohm (1 Ω) : Resistance of a conductor is said to be 1 Ω, if current of 1 A flows through it, when potential difference of 1 V is applied across it.

Cause of Resistance of a conductor:

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

When a conductor is connected across a cell or battery, free electrons flow from negative terminal to positive terminal in conductor. When free electrons flow from one end to another end of the conductor, they collide with the ions in the conductor. This collision offers opposition to the flow of electrons through the conductor.

FACTORS ON WHICH RESISTANCE DEPENDS:

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

1. On length of conductor
2. On area of cross-section
3. Nature of material
4. Temperature.

1. On length of conductor:

The resistance of a conductor is directly proportional to its length l 

R ∝ l

2. On area of cross-section:

The resistance of a conductor is inversely proportional to its area of cross-section

R ∝ 1/a

3. Nature of material: 

Resistance of a material depends weather it is a conductor , Insulator,Semiconductor and alloys.

Conductor: Low

Insulator: Very High

Semiconductor: Nearly High

Alloys: Moderate.

Further different conductors has different value of resistance.
CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

4. Temperature of Material:

Conductors: with the increase in temperature resistance of conductor increases.

Insulators and Semiconductors: Resistance decreases with increase in temperature.

Alloys:Their resistance does not change appreciably with the change in temperature. 

RESISTIVITY(ρ):

we know that 

Resistance of a conductor depends upon:

i)Length: The resistance of a conductor is directly proportional to its length l 

R ∝ l.........(i)

ii) Area of cross-section: The resistance of a conductor is inversely proportional to its area of cross-section

R ∝ 1/a...........(ii)

Now           R ∝ l/a
                 R = ρl/a

ρ(rho) is known as resistivity of the substance.

                 ρ = R(a/l)

If a=1 , l=1 , then ρ=R

Definition: The resistivity of a material is equal to the resistance offered by a wire of unit length and unit cross-sectional area when current flow in parallel sides in perpendicular direction is called resistivity. 

S I unit is Ωm.

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES
NOTE:
1. Resistivity of a conductor depends upon the temperature of the conductor also.Temperature of a conductor increase, its resistivity also increases.
2. Resistivity of an object does not depend upon the dimensions of the objects.
3. It depends upon the nature of the material of the conductor.
4. Alloys have high resistivity and do not oxidize (burn) readily at high temperature, for this reason they are commonly used in electrical heating devices, like electric iron,heater etc.

Resistivity of various materials:

Material	Resistivity (Ω m) At 20 0C		Material	Resistivity (Ω m) At 20 0C
(a)    Conductor			(b) Semi conductors
Silver	1.6 x 10-8		Carbon	3.5 x 10-5
Copper	1.8 x 10-8		Germanium	0.60
Aluminium	2.8 x 10-8		Silicon	2300
Tungsten	5.6 x 10-8
Brass	7 x 10-8		(c) Insulator
Iron	11 x 10-8		Glass	1012
Manganin	44 x 10-8		Wood	1013
Constantan	49 x 10-8		Amber	5 x 1014
Mercury	96 x 10-8		Mica	1011-1015
Nichrome	100 x 10-8		Quartz	7.5 x 1016

Q NCERT:

 A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Ans.

 Diameter of wire (D) = 0.5 mm, = 0.5 x 10^-3m= 5 x 10^-4 m
Resistivity (ρ) = 1.6 x 10^-8 Ω m,
Resistance (R) = 10 Ω.
R = ρ L/A = ρ L/ π(D/2)²= 4 ρ L/ πD²

L= πD² R/4ρ

= 22 x (5 x 10^-4)² x 10/ 7 x 4 x 1.6 x 10^-8 
= 220 x 25 x 10^-8/28 x 1.6 x 10^-8)
= 5500 x 10^-8/ 44.8 x 10^-8
= 5500 / 44.8
= 122.5 m

The resistance is inversely proportional to the cross-section area of wire for given length of given material so, If the diameter is doubled,
Let new diameter is(d) = 2 D 

R^’= ρ L/A= 4ρ L/ πd²= 4ρ L/ π(2D)²

= 4ρ L/ 4πD²= (4ρ L/ πD²)/4 

= R/4 =10/4=2.5 Ω
Resistance of wire will become 1/4th of original wire i.e 2.5Ω

RESISTANCE OF SYSTEM OF RESISTORS

/COMBINATION OF RESISTORS:

The resistors may be connected mainly in two ways : 
(i) in series and (ii) in parallel.

There can be mixed grouping and complex grouping of resistors also. 

Equivalent resistor: A single resistor which draws the same current as the given combination of resistors ( in series or in parallel) when the same potential difference is applied across its end points.

(i) SERIES COMBINATION : 

Two or more resistors are said to be connected in series if they are connected one after the other such that same current flows through them when some potential difference applied across the combination.

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

Let us consider three resistors R₁ , R₂ and R₃ in series and potential difference of V is applied across, current I flows in the circuit.

Resistors in Series

V = V₁ + V₂ + V₃

Using Ohm’s law

I R = I R₁ + I R₂ + I R₃

I R = I ( R₁ +  R₂ +  R₃)

R_s = R₁ +  R₂ +  R₃  

Note: 

1. Equivalent resistance of series combination is equal to the sum of individual resistances.

e g

 R₁ = 1 Ω , R₂ = 2 Ω , R₃ = 3 Ω ,

R_S = ( R₁ +  R₂ +  R₃) = 1 + 2 + 3 = 6 Ω

R_S = 6 Ω  > R₃ = 3 Ω 

2.The total resistance in series circuit is more than the greatest resistance in the circuit.

3. For n resistors in series :

R_s = R₁ +  R₂ +  R₃ +…….+R_n 

(ii) PARALLEL COMBINATION:

Two or more resistors are said to be connected in parallel if all resistors have same potential difference across combination.

Let us consider three resistors R₁ , R₂ and R₃ in parallel and potential difference of V is applied across, current I flows in the circuit.

Resistors in Parallel

I = I₁ + I₂ + I₃ 

Using I=V/R

V/R_P = V/R₁ + V/R₂ + V/R₃

V/R_P = V(1/R₁ + 1/R₂ + 1/R₃ )

1/R_P = (1/R₁ + 1/R₂ + 1/R₃ )

NOTE: 

1. In parallel circuit, equivalent resistance is less than the least resistance in the circuit. e g 

R₁ = 2 Ω , R₂ = 1 Ω , R₃ = 4 Ω ,

1/R_P = ( 1/R₁ +  1/R₂ +  1/R₃) = 1/2 + 1/1 + 1/4 = (2+4+1)/4  = 7/4 Ω

1/R_P = 7/4 Ω

R_P = 4/7 Ω =0.57 Ω

R_P = 4/7 Ω = 0.57 Ω  < R₂ = 1 Ω.

2.

 1/R_P = 1/R₁ + 1/R₂ = R₂ + R₁ / R₁ R₂

OR R_P = R₁ R₂/ R₂ + R₁ 

3.  For n resistors in parallel

1/R_P = (1/R₁ + 1/R₂ + 1/R₃+ ….+ 1/R_n )

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

DISADVANTAGE OF USING SERIES COMBINATION:

1. In series combination if any of device fail to work , the circuit will break and none of the devices will work.

2. Different devices need current according to their resistance to operate efficiently , but if joined in series all have to work on same current.

3. Total resistance in the series circuit increases hence consumption of power also increases.

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

ADVANTAGE OF USING PARALLEL COMBINATION:

1.Voltage across each connecting electrical device is same and device take current as per its resistance.

2. Separate on off switches can be applied on each device.

3.Total resistance in parallel circuit decreases hence a great current may be drawn from the cell.

4.If one device get damaged other can still work.

HEATING EFFECT OF ELECTRIC CURRENT:

The process of heating a conductor, when electric current passes through the conductor is known as heating effect of electric current.
CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

JOULE'S HEATING EFFECT OF ELECTRIC CURRENT:

The heat produced in a given conductor depends upon the square of the current flowing through it, resistance of the conductor and time for which the current flows through it.

H = I²Rt

Consider a conductor AB of resistance R. Let I current flowing through the conductor for time t when potential difference of V volt is applied across the ends of the conductor, then total charge flowing through the conductor is given by,

q = It     (as I=q/t)

From the definition of potential difference ,V = W/q

Where W is the work done in carrying charge q from A to B.

W = Vq                using q = It      

W = VIt                  

W = IRIt              using Ohm's Law V = IR

W = I²Rt

This work done is equal to the heat produced (H)in the conductor.

H = W =  I²Rt 

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

Heat energy due to electric current can be expressed in different ways e.g.

H = VIt =  I²Rt =  V²t/R 


H = VIt  (Using Ohm's Law i e V=IR)

H = IRIt = I²Rt (Using Ohm's Law I= V/R )

H = I²Rt = V²Rt/R² = V²t/R  

Application of Heating Effect of Electric Current:

Electric iron , electric heater , electric oven , geyser and electric fuse etc.

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

ELECTRIC FUSE: 

Electric fuse is a productive device used in series with an electric circuit to save it when abnormal current passes through the circuit.

Fuse is a small piece of wire made up of a material having suitable resistivity and low melting point e g alloy of tin and lead.

Electric fuse are of different rating 1 A , 2 A , 3 A , 5 A , 10 A etc. 

Whenever excessive current flows through the circuit the fuse wire melt down and breaks the circuit. 

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

ELECTRIC POWER (P):

Electric power is defined as the rate of doing electrical work.

Electric power is defined as rate of change of electric energy consumed or dissipated in an electric circuit.

Electric Power = Energy consumed / time

P = E / t

P =  I²Rt / t = I²R

P =  V²t/Rt =  V²/R  

P = V I t /t = V I

 Electric power is also defined as the product of the applied voltage and current flowing through the circuit.

S I unit of Electric Power is watt (W).

1 W = 1 V A

Electric Power is said to be 1 W if 1 A of current flows through an electric circuit, when potential difference of 1 V is applied across it.

Practical unit of power is horse power (h.p.).

1 h.p. = 746 W

RELATION BETWEEN ELECTRIC ENERGY AND ELECTRIC POWER

E = P t 

1 unit = 1kWh = 1000 Wh

1 kWh = 1000 W h = 1000 Js^-1 x 3600 s = 3600000 J = 3.6 x 10⁶ J

1 kWh = 3.6 x 10⁶ J

CBSE CLASS X SCIENCE CHAPTER ELECTRICITY TOPIC RESISTANCE NOTES

CBSE EXTRA QUESTION ON CHAPTER 12 ELECTRICITY