Class X – Electricity (PYQs) – Solutions

Ans. Ω m (Ohm metre)

2. Mention two reasons why tungsten is used for making filaments of electric lamps.

Ans. (i) High resistivity, (ii) High melting point

Ans. Alloys have higher resistivity as compared to pure metals.

Ans. The wire used in the elements of an electric heater has high melting point whereas a fuse wire has low melting point.

Ans. Cell or battery eliminator.

Ans. Its resistance increases.

Ans. It is an alloy with high resistivity and high melting point.

Ans. Resistance decreases, R ∝ l/A, Resistance is directly proportional to length but inversely proportional to area of cross-section.

Ans. (i) Length, (ii) Area of cross section

It is directly proportional to the length and inversely proportional to the area of cross section of a resistor.

Ans. The direction of conventional current is opposite to the direction of flow of electrons.

Ans. Length becomes 4 times whereas area becomes 1/4 times of the original value. Since, R = ρL/A = ρ4L/A/4 = ρ 16L/A =16ρL/A where ρ (rho) is resistivity, L is length and ‘A’ is area of cross section. Resistance will become 16 times of the original value.

Ans. ‘Y’ has more resistance because resistance is directly proportional to the length of the wire.

R₁ = ρ x L/A                                                         (i)

R₂ = ρ x 2L/A/2 = 4ρ x L/A = 4R₁                 (ii)

R₃ = ρ x L/2/2A = 1/ x ρ x L/A = 1/4R₁      (iii)

Hence, the cylinder having length (2L) and area of cross-section (A/2) has the highest resistance followed by the cylinder having length (L) and area of cross section (a), while cylinder with length (L/2) and area of cross-section (2A) has the least resistance. This is because resistance of a substance is directly proportional to the length and is inversely proportional to the area of cross-section.

Thus in figure (b) the resistance is maximum. So, from equations (i), (ii) and (iii), we get R2 > R1 > R3

Ans. R = ρl/A

Ρ = 1.72 x 10^-5 Ω m; r = 1mm = 1 x 10^-3 m; A = πr² = 22/7 x 1 x 10^-6

l = 1 Km = 1000 m

=> R = (1.72 x 10^-8 x 1000 x 7)/ (22 x 1 x 10^-6) = 12.04/22 x 10^-8+3+6 = 120.4/22 = 5.473 Ω

Ans. R = ρl/A

R = 10 Ω, r = 0.01 x 10^-2 m = 1 x 10^-4, A = πr² = 22/7 x 1 x 10^-8

Ρ = 50 x 10^-8 Ω m

l = R x A / ρ = (10 x 22 x 10^-8)/(50 x 10^-8 x 7) = 220/350 = 0.628 m = 62.8 cm

Ans. R = ρl/A

A = (1.6 x 10^-8 Ω m x 250 m)/1 Ω = 4 x 10^-6 m²

If diameter is doubled then area of cross-section becomes four times and the resistance will become one fourth, i.e., 0.25 W.

Ans. (i) Connect the device in the circuit with battery.

(ii) Close the circuit and notice the deflection of the pointer.

(iii) If the deflection is in the opposite direction i.e. below zero, then interchange the terminals.

Ans. (i) The SI unit of resistance is Ohm (W).

(ii) If the current of 1 ampere flows through a wire on applying a potential difference of one volt across it, then the resistance of the wire is said to be 1 Ω.

(iii) I = 200 mA = 200 × 10^–3 A

R = V/I =0.8/0.2 = 4 ohms

Ans. (i) V = 220 V, R = 1200 Ω

I = V/R = 220/1200 = 0.18A (By Ohm’s law: V = IR)

Ans. Now, V = 220 V, R = 100 Ω

I = 220/100 = 2.2A

1mA = 10^–3 A, 1 μA = 10^–6 A

Ans. Resistivity (r) = RA/l

Ans. Its SI unit is Ω m

Ans. Resistivity is a characteristic property of a material that does not depend upon the dimensions of the material whereas resistance depends upon the dimensions of the material.

Ans. Resistivity does not depend on the:

(a) length of conductor, (b) area of cross section of conductor

Resistivity depends on the:

(a) nature of material of conductor (b) temperature of conductor

Ans. Ohm’s Law: It states that ratio of potential difference and current is constant and is equal to the resistance of the conductor at a particular temperature. In other words, the current flowing through a conductor is directly proportional to the potential difference at a constant temperature. The graph between V and I will be a straight line.

Slope = tan θ = ‘R’

Slope = BC/AC = R

Ans. Speaking over it in the morning assembly in school.

• Putting posters over it on the school notice board.

Ans. If we save electricity, it can be used by those villages which do not have electricity.

• It can be used in industries, agriculture and for other useful purposes.

• It improves the national economy because high speed trains, industries, development in villages depends upon electricity.

Ans. Voltage-drop is same across both resistors.

Ans. The glow of bulbs B₂ and B₃ will remain same because they are in parallel connection with bulb B₁.

A₁ shows 1 ampere reading, A₂ shows zero and A3 show 1 ampere reading. ‘A’ will show 2 A reading.

Ans. Circuit in series vs Circuit in Parallel (Disadvantages of series circuit):

(a) If one electrical appliance stops working due to some factor, then all the other appliances also stop working.

(b) All the electrical appliances have only one switch due to which they can’t turned on or off individually.

(c) The appliances do not get the same voltage as of the input power supply.

(d) Overall resistance increases due to which the current from the input power supply becomes low.

Ans. Two resistances of 2 Ω are in series R_s = R₁ + R₂ = 2 + 2 = 4 Ω

The third resistance of 4 Ω is parallel to R_s = 1/R_s + 1/R₃ = ¼ + ¼ = 2/4

R_p = 2 Ω

Two 9 ohm resistors in parallel combination are connected to one 9-ohm resistor in series
1/R_p = 1/9 + 1/9 = 2/9
R_p = 9.2 Ω
R = 9 Ω + 9/2 Ω = 13.5 Ω            
ii) Two 9 ohm resistors in series connected are connected to one 9-ohm resistor in parallel;
R_s = 9 Ω + 9 Ω = 18 Ω
1/R = 1/18 + 1/9 = 3/18
R = 6 Ω

Ans. (i) R_s = R₁ + R₂ = 10 Ω + 20 Ω = 30 Ω

I = V/R = 3V/30 Ω= 0.1 A

(ii) V = I × R = 0.1 × 10 = 1 V

Ans. Resistors in Series: When resistors are joined end to end, it is called series combination of resistances.

In series combination of resistors, the current remains the same through each resistor. Therefore, the value of current in the ammeter remains the same, independent of its position in the electric circuit.

In the above circuit, V is equal to the sum of V₁, V₂, V₃, that is the total potential difference (V) across a combination of resistors in series is equal to the sum of potential differences across the individual resistors.

V = V₁ + V₂ + V₃

The current through each resistor is I.

V = IR

V₁ = IR₁ V₂ = IR₂ V₃ = IR₃

As V = V₁ + V₂ + V₃ ⇒  IR = IR₁ + IR₂ + IR₃

⇒ R_S = R₁ + R₂ + R₃

We can conclude that when several resistors are joined in series, then equivalent resistance of the combination is equal to the sum of their individual resistances and is thus greater than any of the individual resistance.

Resistors in Parallel:

When the resistances, says, R₁, R₂, R₃ are connected in in parallel combination as shown in the diagram, then the potential difference remains the same across each resistor in parallel combination of resistors. In parallel combination the current is distributed in different resistances in different branches.

As, I = I₁ + I₂ + I₃

According to Ohm’s Law, I = V/R_p

⇒ I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃

 I = I₁ + I₂ + I₃

• V/R_p = V/R₁ + V/R₂ + V/R₃
• 1/R_p = 1/R₁ + 1/R₂ + 1/R₃

Ans. Resistance across parallel combination,

1/R_p = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅

Resultant resistance of 5 Ω, 20 Ω, 15 Ω, 10 Ω, and 20 Ω resistances is given as:

1/R_p = 1/5 + 1/20 + 1/15 + 1/10 + 1/20 => 1/R_p = (12 + 3 + 4 + 6 + 3)/60

1/R_p = 28/60 => R_p = 60/28 Ω => R_p = 2.14 Ω

(b) V= 6 V, Resistance = 60/28 Ω

I = V/R = (6 x 28)/60 = I => 2.8A

Ans. Alternate current

Ans. H = VIt

Ans. Alloys have high resistivity/high melting point/alloys do not oxidise (or burn readily at high temperatures).

Ans. P = V²/R => R = V²/P = (250 x 250)/100 = 625 Ω

Ans. I = P/V = (1.1 x 1000)/220 = 1100/220 = 5A

The rating of fuse that can be used is more than 5 A. It is also called Board of Trade Unit (BOTU)

Ans. R_{100 Ω} = V²/P = (220 x 220)/100; R_{60 Ω} = V²/P = (220 x 220)/60

I = V/R₁₀₀ = (220 x 100)/(220 x 220) = 100/220 = 0.45 A; I = V/R₆₀ = (220 x 60)/(220 x 220) = 6/22 = 0.27 A;

Total Current = 0.45 A + 0.27 A = 0.72 A

Ans. The current whose direction gets reversed after every half cycle is called alternating current or A.C. There is no change in the direction of D.C. D.C. is a unidirectional current. The most important advantage of using A.C. over D.C. is that in the A.C. mode electric power can be transmitted over long distances at a high voltage and low current with very little loss of power.

Ans. Here P = 2 kW = 2000 W, V = 220 Volt

P = VI, so the current, I = P/V = 2000/220 = 9.09 A

As the current is 9.09 A below the rating of fuse, therefore the fuse will withstand, i.e. it will not blow off when A.C. is on.

Ans. It is in opposite direction offered to the flow of electric current.

(i) No effect on resistance, low current, hence no appreciable rise in temperature so there no change in resistance.

(ii) Heavy current for 30 seconds may increase the temperature so resistance will increase.

Ans. P = 2 kW, t = 2 h

E =P × t = 2 × 2 h = 4 kW h

Ans. Total energy consumed per month = 4 kW h × 30 = 120 kW h

Total cost = 120 × 3 = ₹360.

Ans. R = V²/P = (220 V x 220 V)/100 W = 484 Ω

P = V²/R = (110 x 110)/484 = 25 W

Ans.

Ans. I = (N x P)/V => 5 A = (N x 40)/220 => N = (220 x 5)/40 = 110/4 = 27 lamps

Ans. The lamps are in parallel.

Ans. Advantages: If one lamp is faulty, it will not affect the working of the other lamps. They will also be using the full potential of the battery as they are connected in parallel.

Ans. The lamp with the highest power will glow the brightest.

P = VI

In this case, all the bulbs have the same voltage. But lamp C has the highest current.

Hence, for lamp C, P = 5 × 60 Watt = 300 W. (the maximum).

Ans. The total current in the circuit = (3 + 4 + 5 + 3) A = 15A

The Voltage = 60V

R = 60/15 W = 4 W

V = IR and hence R = V/I

Ans. Resistivity will not change as it depends on the nature of the material of the conductor.

Ans. The length of each part becomes L/4. While ρ, A remain constant. 

R = ρL/A.

Resistance of each part = R_part = (ρL/4)/A = R/4.

(a) In parallel connection

In series connection,

Ans. P = V²/R.

If R_eqv is less, power consumed will be more.

In the given case, R_eqv is lesser in the parallel connection and the power consumed will be more.