## Electricity - (Holiday Assignment)

Q1: Two metallic wires A and B of same material are connected in parallel. Wire A has length l and radius r and wire B has length 2l and radius 2r. Compare the ratio of the total resistance of parallel combination and the resistance of wire A.

Wire A:
Length = l
Area   = πr²
R₁   = ρ  x l /(πr²)

Wire B:
Length = 2l
Area = π(2r)² = 4πr²
R₂   = ρ  x 2l /(4πr²) = ρ  x l /(2πr²)

Let Rₚ = effective resistance of parallel combination
1/Rₚ = 1/R₁ + 1/R₂

1/Rₚ = (πr²)/ρl + (2πr²)/ρl

Rₚ = ρl/(3πr²)

Thus,
Rₚ/R₁ = [ρl/(3πr²)]/ [ρl/(πr²)] = 1/3
Rₚ : R₁ = 1 : 3

Q2: A wire is stretched so that its length becomes 6/5 times its original length. If its original resistance is 25 ohm, find its new resistance.

Let l₁ is the length and A₁ is cross section area.
R₁   = ρ  x l₁ /A₁ = 25 Ώ

When R₁  wire is strteched 6/5 times (l₂ = 6l₁/5), its volume remains constant.
i.e. Old Volume = new Volume
A₁  x l₁ = A₂ x l₂ = A₂ x  6l₁/5
A₂ = 5A₁/6

R₂   = ρ  x l₂ /A₂
R₂   = ρ x (6l₁/5) / (5A₁/6)
= (36/25) ρ  x l₁ /A₁
= (36/25) x 25  Ώ
= 36 Ώ

Q3: List two reasons why nichrome is used for making heating element of electrical appliances.

Nichrome being an alloy has high resistivity and higher resistance.
Alloys do not oxidise (burn) readily at high temperatures.
Nichrome wire has high melting point. Thus it produces heat, when current is passed through it.

Q4: Why are Connecting wires in a circuit made of copper and aluminium?

Copper and aluminium are good conductors of electricity.
They have low electrical resistance.
As they are malleable and ductile they can be drawn into thin wires.

Hence connecting wires in a circuit is made of copper or aluminium.

Q5: Two resistances when connected in series,their resultant resistance is 10 ohm, but when they are connected in parallel their resistance is 2.4 ohm. Find the value of each resistance separately.

Answer: Let R₁ and R₂  be two resistors.
When connected in series,
R₁ + R₂ = 10 Ώ
R₂ = 10 - R₁

When connected in parallel
1/Rₚ = 1/R₁ + 1/R₂
1/2.4 = (R₁ + R₂)/(R₁R₂)

2.4 = (R₁R₂)/ (R₁ + R₂)

2.4 = (R₁R₂) / 10
R₁R₂ = 24
R₁(10 - R₁) = 24

10R₁ - R₁² = 24

⇒ R₁² - 10R₁ + 24 = 0
R₁² - 6R₁ - 4R₁ + 24 = 0
R₁(R₁ - 6) - 4(R₁ - 6) = 0
(R₁ - 6)(R₁ - 4) = 0
⇒ R₁ = 6 Ώ  and R₁ = 4Ώ
Thus  R₁ = 6 Ώ  and R₂ = 4Ώ
Or  R₁ = 4 Ώ  and R₂ = 6Ώ

Q6: State Ohm's law.

Answer: At constant temperature, the potential difference between the ends of a conductor is directly proportional to electric current passing through it.

Q7: Why do we consider tungsten as suitable material for making the filament of a bulb?

Answer: The filament of an electric bulb is usually made of tungsten, because of its higher resistivity value (5.56 ×10⁻⁸  Ωm) and its high melting point (3422°c).  