Question

Two bulbs rate 200W at 220V and 400W at 220V are connected in parallel to a 220V line. What total current is drawn by them​

Answer 1

Answer :-

Total current drawn by the bulbs is 2.72 Amperes .

Explanation :-

We have :-

→ Votage of 1st and 2nd bulb (V) = 220 V

→ Power of 1st bulb (P₁) = 200 W

→ Power of 2nd bulb (P₂) = 400 W

→ Voltage of the line (V') = 220 V

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For the 1st bulb :-

⇒ R₁ = V²/P₁

⇒ R₁ = (220)²/200

⇒ R₁ = 48400/200

⇒ R₁ = 242 Ω

For 2nd bulb :-

⇒ R₂ = V²/P₂

⇒ R₂ = (220)²/400

⇒ R₂ = 48400/400

⇒ R₂ = 141 Ω

The two bulbs are connected in parallel. So, equivalent resistance will be :-

1/R = 1/R₁ + 1/R₂

⇒ 1/R = 1/242 + 1/141

⇒ 1/R = 3/242

⇒ R = 242/3

⇒ R = 80.67 Ω

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Now, let's calculate the total current drawn using Ohm's Law .

V' = IR

⇒ 220 = I × 80.67

⇒ I = 220/80.67

⇒ I = 2.72 A

Answer 2

Given :-

Two bulbs rate 200W at 220V and 400W at 220V are connected in parallel to a 220V line

To Find :-

Total current

Solution :-

We know that

Case 1

P = V²/R

200 = (220)²/R

200 = 48400/R

48400/200 = R

484/2 = R

242 = R

Case 2

P = V²/R

400 = (220)²/R

400 = 48400/R

48400/400 = R

484/4 = R

121 = R

Now,

1/Rn = 1/R1 + 1/R2 = R1 × R2/R1 + R2

Rn = 242 × 121/(242 + 121)

Rn = 29282/363

Rn = 80.66 Ω

Now

I = V/R

I = 220/80.66

I = 22000/8066

I = 2.72 A

Therefore, Current drawn is 2.72 A