Class 10 Science Chapter 11 Board Questions

In series.

Resistance decreases as R ∝ 1/A

Length becomes one-fourth of the original length and area of cross-section becomes four times that of original.
i.e., I₂ = 1/4 I₁ and A₂ = 4A₁
R₂/R₁ = I₂/I₁ × A₁/A₂ = 1/4×1/4 = 1/16
R₂ = 1/16 R₁
So, new resistance is (1/16)th of original resistance.

In series same current flows through each resistor, SO ratio of current 1:1.

Resistance of each side of a square = 20/4 = 5∩
1/Req = 1/(RAB + RBC) + 1/(RAD + RDC)
= (1 )/(5 + 5) + 1/(5 + 5)
= 1/10 + 1/10 = 1/5

If one stops working due to some reason, other will also stop working.

P = (V2 )/(R )
P ∝ 1/4 R ∝ 1/P
So, 60 W bulb has a higher resistance.

If four resistors each o each 1 ∩ are connected in parallel the effective resistance will be 0.25∩.

Resistivity of a given conductor depends only on its material and the temperature. Resistivity does not depend on the dimensions of the conductor.

Time rate of flow of charge through any cross-section of a conductor is called electric current.

An ampere (A) which is defined as the rate of flow of 1 coulomb of charge per second.

In an electric circuit, the direction of conventional current is taken as opposite to the direction of flow of electrons.

Potential difference between two points A and B in an electric field is defined as the amount of work done in order to move unit positive charge from point B to point A. thus,
Vᴀ – Vʙ = (Wᴀʙ)/q

Amount of work done to bring 1C charge from point B to point A in the electric field is 1 joule.

Potential difference between two points of an electric circuit is said to be 1 Volt, when a work of 1 J is to be done for moving a charge of 1 C between these two points.

Electrical resistance.

Resistance of a conductor is the measure of opposition offered by it for the flow of electrical charge through it. SI unit of resistance is ohm.

Resistance of a wire is said to be 1∩ if for flow of 1A current through the wire one has to maintain a potential difference of 1volt across its ends.

Current
I = (Potentail difference V)/(Resistance R)
= 1.5V/(30∩)
= 0.05A

Silver.

A straight inclined line passing through the original because for a metallic wire V∝I.

The current flowering through the conductor is reduced to one half of its previous value in accordance with the relation I = V/R.

The reading of ammeter (current flowering in the circuit) is reduced to one half of its previous value because I = V/R.

There is no change because resistivity of a material depends only on its nature and is independent of its dimensions.

The electrical resistivity of a material is the resistance offered by unit cube of that material between its opposite faces. Its SI unit is ohm-meter.

Resistance of a conductor (i) a directly proportional to its length, (ii) inversely proportional to its cross-section areas and depend on the material of the conductor. Resistance also dependence on the temperature.

Resistance of a conductor (i) a directly proportional to its length, (ii) inversely proportional to its cross-section areas and depend on the material of the conductor. Resistance also dependence on the temperature.

For maximum current flow the net resistance of circuit must be least possible. Hence resistor R₁ and R₂ should be connected parallel.

Potential difference
V = RI
= 18 × 5
= 90 V.

In parallel because then 1/R = 1/(6 ) + 1/3 = (1+2)/6 = 1/2.
Hence, R = 2 Ω.

The resistance of conductor is reduced to one-half of its previous value because resistance R ∝ Length L.

On increasing the temperature the resistance of given conductor also increases.

An ammeter.

In series of electric current.

(i) Ammeter
(ii) Voltmeter

In series arrangement of resistance same current flows through all the resistors.

In parallel arrangement current flowing through different resistors are in the inverse ratio of there resistance. Hence current passing through 30 Ω resistor is double of current passing through 60 Ω resistor.

Power = Potential difference × Current.

When electric current is pass through a resistor electrical energy is dissipated and appear as heat energy. This is known as the heating effect of electric current.

3.6 × 10^6 J = 1 kWh.

40 W lamp has a higher electrical resistance because R = V²/P.

Energy consumed
E = Pt = 1 W × 1 s
= 1 J.

Toaster has a greater resistance because its power is less and R=(V²)/P

Electric iron, electric toaster, electrical oven etc.

Nichrome is an alloy of high resistivity and high melting point and does not oxidise.

Given V: 220V, R₁ =1200Ω, I₁ =?
R₂ = 100Ω, I₂ =?
Using Ohms law,
V= I₁R₂
I₁ = V/(R₁) = 220/1200 = 0.18A
And, I₂ = V/(R₂) = 220/100 = 2.2A

Resistance of each part = R/3 = 9/3 = 3Ω
R₁ = R₂ = R₃ = 3Ω
In parallel combination,
1/(Rₚ) = 1/(R₁) + 1/(R₂) + 1/(R₃) = 1/3 + 1/3 + 1/3 = 3/3 = 1
Rₚ = 1Ω

Given: P = 750 W, V = 220V
(i) P = VI
750 = 220 × I
I = 750/220 = 3.40A
(ii) P = (V²)/R
R = (V²)/R = (220²)/R , R= 64.53Ω

Given: q= 1.6 × 10⁻¹⁹ C, I= 1A, n=?, t=1s
We know, q=It and q =ne
ne = It
n = (It )/(e ) = (1×1)/(1.6×10⁻¹⁹)

Fuse wire is a safety devices connected in series with the live wire of circuit. It has high resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But in disconnects the entire circuit from the electrical supply. But in case if we use a larger rating instead of a defined rating, then it will not protect the circuit as high current will easily pass through it and it will not melt.

Given P₁ = 24 W, V₁ = 12V, P^2 =?, V^2 = 6V
Using P = (V^2)/R
(P₁)/(P^2) = (V₁)/(V^2)
P^2 = [ (V^2)/(V₁)] × P₁
[6/12]^2 × 24 = 1/4× 24 = 6W

For wire A:
I₁ = l, r₁ = r, p₁ = p
So Rᴀ = p₁ (l₁)/(A₁) = p₁(l₁)/(πr^2) =p₁/(πr^2)
For wire B:
l₂ =2l, r₂ =2r, p₂ =p
Rʙ = P₂(l₂)/(A₂) = p₂(l₁)/(πr^2) = (p(2l))/(π(2r)^2)
1/2 = {pl/(πr^2)} = (Rᴀ)/2
In parallel combination of Rᴀ and Rʙ
1/(Rₚ) = 1/(Rᴀ) + 1/(Rʙ) = 1/(Rᴀ) + 1/(Rᴀl₂) = 3/(Rᴀ)
(Rₚ)/(Rᴀ) = 1/3 = 1:3

Using R= pl/A
We have, (R₁)/(R₂) = (I₁)/(A₁). (A₂)/(I₂)
Given I₂ =2l₁
Volume of material will be conserved. So A₁l₁ = A₂l₂
(A₁)/(A₂) = (l₂)/(l₁) = 2
(R₁)/(R₂) = (l₁)/(l₂). (l₂)/(l₁) = (l₂/l₁ )² = 1/4
R₂ = 4R₁ = 80Ω

(i) R= (V²)/P = (220²)/25 = (220×220)/25 = 1936 Ω
(ii) Energy consumed = Power × Time
= 25 × 10 = 250 Wh = 0.25 kWh

(i) Copper as it has the lesser resistivity.
(ii) Manganin as it has the comparatively higher resistivity, less oxidation even at high temperature.

(a) Electric current in a circuit is defined as the time rate of flow of electric charge through any cross-section and its direction is opposite to that of flow of electrons. Hence in present case
Electric current I = q/t = ne/1 = ne
As electron are flowing from east to west, the direction of electric current is from west to east.
(b) Here current I= 1A, time t= 1s and charge on each electron e = 1.6 × 10⁻¹⁹ C.
Hence number of electrons flowing n= (It )/e = (1 ×1)/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸

(a) The term ‘volt’ is the SI unit of potential difference. Potential difference is said to be 1 volt if one joules work is to be done to carry 1 coulomb charge from one point to another.
(b) The relation between work (W), charge (Q) and potential difference (v) for an electric circuit is
V = (W )/Q
Given that W = 100J and Q = 20 C
Potential difference V = (W )/Q = (100J )/(20 C ) = 5V

(a) Electric charges can move in a conductor when potential differences is maintained across its ends. An electric cell is used to maintain potential difference across the conductor and thus can be used for flow of charges.
(b) Here I = 2A, time t = 1 minutes = 60s and V = 3V
Total charge flowing q = It = 2 × 60 = 120C
Work done W = qV =120 × 3 = 360 J

George Simon ohm established a relationship between the electric current flowing through a conductor and the potential difference across its ends due to which current flows. According to Ohms law temperature remaining constant the current passing through a conductor is directly proportional to the potential difference across its ends
V ∝ I or V = IR
Here constant R is known as the resistance of given conductor. For a given conductor its resistance is constant at a given temperature. XY is a resistance wire, An ammeter and V a voltmeter. A battery of four cells is being used as a current source and K is a plug key. Initially use one cell only. Put plug in key K and note current and voltage by nothing ammeter and voltmeter reading respectively. Let those be I₁ and V₁. Then connect two cells in the circuit and note current I₂ and potential difference V₂ across the resistance. Similarly take reading with three cells and four cells in the circuit. From our observations, we find that
(V₁)/V₂ = (V₂ )/I₂ = (V₃ )/I₃ = (V₄ )/I₄ = a constant = R.

(a) The resistance of a conductor is a property of the conductor which affects the flow of current through it on maintaining a potential difference across its ends.
Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm if a potential difference of 1V is to be applied across its ends for maintaining flow of 1A current.
(b) If given resistance wire is replaced by another thicker wire of same material and same length then cross-section area wire is increased and consequently its resistance decreases. So the current flowing in the circuit increases.

(a) The resistance of a cylindrical conductor a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depend on nature of material.
R ∝ (L)/A
Here p is known as the resistivity of given material. It is defined as the resistance offered by a unit cube of given material when current flows perpendicular to the opposites faces.
(b) The resistivity of the conductor remains unchanged.

To demonstrate dependence of resistance on length, cross-section area and material of the conductor we complete an electric circuit.
(i) Effect of length: Take the resistance wire number 1 of length l and connect in the circuit. Put plug in key K and note ammeter reading I₁. Now instead of wire number 1, connected wire number 2 of double length but of same material and same thickness. As before note ammeter reading I₂. Experimentally it is observed I₂ = (I₁)/2 It shows the resistance of 2nd wire is double of 1st wire thus, we conclude that
(R₂)/R₁ = (l₂)/I₁ = (2l )/l = 2
(ii) Effect of cross-section area: Replace the wire by resistance wire number 3 of length l but thickness more than wire number 1. Put plug in key and note ammeter reading I₃. It is more than I₁. It means for a thicker wire the resistance is less. Extract calculation shows that resistance is inversely proportional to cross-section area of the conductor.
Thus R∝ (1 )/A
(iii) Effect of material: Now take a wire number of 4 of some other material but of length l and same thickness as wire number 1. Connect the wire in the circuit and again note ammeter reading I₄ is difference from I₁. It shows that resistance depends on material of the conductor.

(a) The resistance R of a conductor, the shape of a cylinder, of length l and area of cross-section A is given as:
R = pl/A
Where p is a constant which is known as the electrical resistivity of the material of conductor.
Thus resistivity p= (RA )/(l )
SI unit of resistivity p shall be (ohm × m² )/m = ohm m= ∩m
(b) Here l = 5m, R = 100 Ω A = 3 × 10^7m²
Resistivity of the material of wire p = (RA )/(l ) = (100 ×3×10⁻⁷)/(5 )
= 6.0 × 10⁻⁶ ohm m

In series combination, the same current flows in all the resistance but the potential difference across each of the resistance is different.
According to Ohm’s law we have
V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
If the total potential difference between A and B is V, then
V = V₁ + V₂ + V₃
= IR₁ + IR₂ + IR₃
= I(R₁ + R₂ + R₃)
Let the equivalent resistance be R then
V = IR
IR = I(R₁ + R₂ + R₃)
R = R₁ + R₂ + R₃.

In parallel combination of three resistance R₁, R₂ and R₃ the current in each of the resistance is different. If I is the current drawn from the cell then it is divide into branches I₁, I₂ and I₃. Thus,
I = I₁ + I₂ + I₃.
Thus from Ohm’s law I₁ = (V )/R₁, I₂ = (V )/R₂, I₃ =(V )/R₃
If R is the equivalent resistance then,
I= (V )/R
(V)/R = (V)/R₁ + (V)/R₂ + (V)/R₃
(1)/R = (1)/R₁ + (1)/R₂ + (1)/R₃

(a) Here power of bulb P = 40W and voltage V= 220V
Current drawn by the bulb I = (P)/V = (40)/220 = (2)/11A
And resistance of the bulb R = (V)/I = (220)/((2/11)) = (220×11)/2 = 1210Ω
(b) On taking another bulb power P= 25V and voltage V= 220V. there is a change in the value of current and resistance because their values depend on the power of the bulb.
New Current I’ = (P’)/V = (25)/220 = (5)/44A
And new resistance R’ = (V)/I’ = (220)/((5/44)) = 1936Ω

Here power P = 2kW, time of use t= 2h daily
Total time of use in the month of June t = 2 × 30 = 60h
(i) Energy consumed E = P × t = 2kW × 60h = 120kWh = 120 units
E = 120 × 3.6 × 10⁶ J = 4.32 × 10^8J
(ii) Cost ₹ 3.00 per unit = 3.6 × 120 = ₹ 360

(a) When heating is at maximum rate the power rating P = 840W and supply voltage V= 220V
Current flowing I= (P)/V = (840W)/220V = 3.82A
and resistance of electric iron R= (V )/I = (220 V)/3.82A =57.6Ω
(b) When heating is at maximum rate the power rating P = 360W
Current flowing I = (P)/V = (360W )/220V = 1.64A
And resistance of electric iron R = (V )/I = (220V )/1.64A = 134.2Ω

As here V= and I=5A, hence resistance of the electric heater
R = (V)/I = (110)/5 = 22∩
When potential difference is increased to V’ = 220V then
(a) the current I’ = (V’)/R = (220)/22 = 10A.
(b) the power P’ = V’I’ = 220×10 = 2200W = 2.2Kw

The current drawn by a 40W, 220V electric lamp I₁ = (P₁)/V = (40)/220 = (2)/11A
As the electric line of rating 220V, 5A hence we can connect n lamps in parallel where
n= (I)/I₁ = (5)/((2/11)) = 27.5
Thus we can safely have connected 27 lamps of 40W, 220V rating to a 220V, 5A line.

Here charge transferred Q = 90000C, potential difference between the terminals of battery V= 40 and time t = 1h =3600s.
Current I = (Q)/t (9000)/3600 = 25A
Amount of heat generated H = VIt = 40×25 × 3600 = 3600000J = 3.6×10-^6
And power expanded P = (H)/t = VI = 40 × 25 = 1000W

Here R₁ = R₂ = 10∩ and V= 6 volt
(i) In series net resistance Rₛ = R₁ + R₂= 10 + 10 + 20∩
And power Pₛ = (V²)/Rₛ = (6²)/20 = 1.8W
(ii) In parallel net resistance (1)/Rₚ
= (1)/R₁ + (1)/R₂ = (1)/10 + (1)/10 = (1)/5
and power Pₚ = (V²)/Rₚ = (6²)/5 =7.2W
(Pₛ)/Pₚ = (1.8)/7.2 = (1)/4

Here rating of torch bulb is V= 5V and I = 500 mA = 0.5A
(i) Power of torch bulb P = VI = 5 × 0.5 = 2.5 W
(ii) Resistance of torch bulb R = (V)/I = (5)/0.5 = 10∩
(iii) Energy consumed by bulb in 4 hours
E = Pt = 2.5 × 4 = 10Wh.

Amount of work done in carrying a charge Q through a potential difference V is
W = QV
Q = It
As power is defined as the rate of doing work hence
Power P = W/t = (VIt )/t = VI.
If R be the value of resistance of the conductor, then V= RI and hence
P = VI = (RI)I = I²R
Again P= VI = V(V/R) = V²/R
Thus in general we can say that electric power is given by
P = VI = I²R = V²/R

Let two resistors of value R each are first joined in series when equivalent resistance
Rₛ = R + R = 2R
On joining these very resistors in parallel let equivalent resistance be Rₚ where
1/Rₚ = 1/R + 1/R = 2/R
Rₛ/Rₚ = 2R/(R/2) = 4/1

(a) According to Ohm’s law, physical conditions like temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference maintained across the ends of the conductor i.e.,
V∝ I or V = IR
Hence, R is known as the resistance of given conductor. It is a property of the conductor which affects the flow of current. For a given conductor value of resistance R is a constant at a given temperature.
(b) As rating of bulb is 220V, 100 W, hence P = V1 = V²/R
Resistance of bulb filament R=V²/P = (220 ×220)/100 = 484 Ω.
When the bulb is operated at a voltage V’=110V, the power consumed is
P= V²/R = (110 × 110)/484 = 25.W]
The power will be New power = 25 W

(a) Electrical resistance of a conductor may be considered as a measure of the opposition offered by it for flow of electric charge through it.
Mathematically, resistance R = V/I ((potential difference))/((current))
SI unit of electrical resistance is ohm (Ω), where 1Ω =1V/1A
(b) At a given temperature resistance of a conductor depends on its (i) length L, (ii) cross-section area A, and (iii) nature of the material of conductor. It is found that R∝ L and R∝ 1/A
Mathematically, ρL/A
Where ρ is a constant known as the resistivity of the material of conductor. Its value depends only on the nature of material of the conductor and the temperature and is independent of the dimensions (i.e. length and cross-section area) of the conductor.
The resistivity of a given material is defined as the resistance offered by a cube of that material of side 1m, when current flows perpendicular to the opposite faces. Its unit is ohm-meter (Ωm).

Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor. If Q charge passes through a section of a conductor in time t then current I= Q/t
SI unit of electric current is an ampere (A). current is said to be one ampere. If rate of flow of electric charge through a cross section of conductor be 1 coulomb per second. Direction of conventional current is taken as the direction of flow of positive charges or opposite to the charges of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B. Here current I = 1A time t= 1s and charged on electron e = 1.6 × 10⁻¹⁹C. let one n electron flow through a section of conductor so that charge passing through the section is Q= ne.
I= Q/t = (ne )/t , n = (It )/e (1×1 )/(1.6 × (10)⁻¹⁹) ) = 6.25 × 10¹⁸

In parallel combination of three resistance is R₁, R₂ and R₃ the current in each of the resistance is different. If I is the current drawn from the cell then it divide into branches I₁, I₂ and I₃. Thus I = I₁ + I₂ + I₃
The potential difference across each of these resistance is same.
Thus from Ohm’s law I₁ = V/R₁ , I₂ = V/R₂ , I₃= V/R₃
I= V/R
V/R = V/R₁ + V/R₂ + V/R₃
1/R = 1/R₁ + 1/R₂ + 1/R₃

As resistance of entire wire bent in the form of a circle is 8Ω, hence resistance of each half part (semi-circle) will be 8/2 = 4Ω.
Now, these two resistances of 4∩ each are joined in parallel between points A and B, hence effective resistance of the circuit R is given as
1/R = 1/4 + 1/4 = 1/2 = R = 2Ω
The ammeter reading= current flowing in the circuit I= V/R = 2V/(2Ω) = 1A

(a) In series combination, the same current flows in all the resistance but the potential difference across each of the resistance is different.
According to Ohm’s law we have
V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
If the total potential difference between A and B is V, then
V = V₁ + V₂ + V₃
= IR₁ + IR₂ + IR₃
= I(R₁ + R₂ + R₃)
Let the equivalent resistance be R then
V = IR
IR = I(R₁ + R₂ + R₃)
R = R₁ + R₂ + R₃.
(b) Here R₁ = 2Ω, R₂ = 3Ω, R₃ = 6Ω
The equivalent resistance R in parallel combination of resistor will be
1/R = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = 1/1 R = 1Ω

(a) (i) In series arrangement equivalent resistance Rₛ = R₁ + R₂ + R₃
(ii) In parallel arrangement equivalent resistance Rₚ is given as:
1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃
(b) Here R₁ = R₂ = 12Ω and V= 3V.
For minimum resistance two resistor must be connected in parallel so that
1/Rₚ = 1/R₁ + 1/R₂ = 1/12 + 1/12 = 1/6 R= 6Ω
Hence power Pₚ = V²/Rₚ = ((3)²)/6 = 24Ω
For maximum resistance two resistance must be connected in series so that
Rₛ = R₁ + R₂ = 12 + 12 = 24Ω
So the power Pₛ = V²/Rₛ = (3)²/24 = 0.375W
Pₚ/(Pₛ ) = (1.5 )/(0.375 ) = 4/1

Series combination of resistor: We take three resistor R₁, R₂ and R₃ and join them in series between the points X and Y in an electric circuit.
(a) Plug the key and note the ammeter reading Then change the position of ammeter to anywhere in between the resistor and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement same current flow through each resistor.
(b) Inert a voltmeter across the ends X and Y of the series combination of resistors. Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.
Take out plug from key K and disconnected the voltmeter. Now inert the voltmeter across the ends of first resistor R₁. Plug the key and note the voltmeter reading V₁. Similarly measures the potential difference across the other two resistors R₂ and R₃ separately. Let these potential differences be V₂ and V₃ respectively. Experimentally we find that V= V₁ + V₂ + V₃ It shows that in series arrangement of resistors total potential difference equal to the sum of potential differences is equal to the sum of potential differences across individual resistors.

Here voltage of battery V = 6V resistance of electric lamp R₁ = 20 Ω and resistance of conductor R₂ = 4Ω
(a) Since R₁ and R₂ are connected in series the total resistance of the circuit.
R= R₁ + R₂ = 20 + 4 = 24Ω
(b) The current through the circuit I = V/R = (6 )/24 =0.25A
(c) (i) Potential differences across the electric lamp V₁ = IR₁ = 0.25× 20 = 5V
(ii) Potential differences across the conductor V₂ = IR₂ = 0.25× 4 = 1V
(d) Power of the lamp P= I²R₁ = (O.25)² × 20 = 1.25 W