## 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.**

Answer:

Wire A:

Length = l

Radius = r

Area = πr²

R₁ = ρ x l /(πr²)

Wire B:

Length = 2l

Radius = 2r

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.**

Answer:

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.**

Answer:

① Nichrome being an alloy has high resistivity and higher resistance.

② Alloys do not oxidise (burn) readily at high temperatures.

③ 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?**

Answer:

① Copper and aluminium are good conductors of electricity.

② They have low electrical resistance.

② 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).