Question

Two heaters are marked 200V, 300W
and 200V, 600W. If the heaters are
connected in series and the combination
connected to a 2001 d.c. supply, find
the current in the circuit.​

Answer 1

Answer:

• The current in the circuit = 1 Ampere

Explanation:

Given that:

• The heaters are connected in series combination.
• Two heaters are marked 200V, 300W and 200V, 600W.

To Find:

• The current in the circuit.

Formula used:

• R = V²/P
• I = V/R
• I = V/(R₁ + R₂)

Where,

• R = Resistance
• V = Potential difference
• P = Power
• R₁ = Resistance of first heater
• R₂ = Resistance of second heater

Finding the resistance of heaters:

We have.

• V = 200V
• P₁ = 300W
• P₂ = 600W

⟶ R₁ = V²/P₁

⟶ R₁ = (200)²/300

⟶ R₁ = 40000/300

⟶ R₁ = 400/3

⟶ R₂ = V²/P₂

⟶ R₂ = (200)²/600

⟶ R₂ = 40000/600

⟶ R₂ = 200/3

Finding current in the circuit:

We have.

• V = 200V
• R₁ = 400/3 Ω
• R₂ = 200/3 Ω

⟶ I = V/(R₁ + R₂)

⟶ I = 200/{400/3 + 200/3}

⟶ I = 200/{(400 + 200)/3}

⟶ I = (200 × 3)/600

⟶ I = 600/600

⟶ I = 1

∴ Current in the circuit = 1 Ampere

Answer 2

Answer:

• 1 Ampere

Step-by-step explanation:

Given

• Two heaters are marked 200V, 300W  and 200V, 600W.
• They are connected in series combination.
• Potential difference offered is 200 V.

To find

• Current in the circuit.

Solution

⇒ Resistance of heaters :

• R = V²/P

⇒ Where :

• R - Resistance
• V - Potential difference
• P - Power

⇒ Substituting with P₁ (300) :

• R₁ = (200)²/300
• R₁ = 400/3

⇒ Substituting with P₂ (600) :

• R₂ = (200)²/600
• R₂ = 200/3

∴ The resistance of heaters is 400/3 Ω and 200/3 Ω.

⇒ Current in circuit :

• I = V/(R)
• I = V/(R₁ + R₂)

⇒ Where :

• I - Current
• V - Potential difference
• R₁ - Resistance 1
• R₂ - Resistance 2

⇒ Substituting :

• I = 200/[400/3 + 200/3]
• I = (200 × 3)/600
• I = 600/600
• I = 1

∴ The current in the circuit is 1 Ampere.