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:

2.72 ampere is the required answer.

Explanation:

Given:-

• Power of 1st bulb ,P = 200W
• Power of 2nd bulb ,P' = 400W
• Potential Difference at each bulb ,V = 220v

To Find:-

• Total Current drawn by these bulb

Solution:-

According to the Question

Firstly we calculate the resistance of each bulb

As we know the relationship between Power Potential Difference and Resistance

• P = V²/R

where,

P denote Power

V denote potential difference

R denote Resistance

Resistance of 1st bulb

➻ 200 = 220²/R

➻ 200 × R = 48400

➻ R = 48400/200

➻ R = 242Ω

Resistance of 2nd bulb

➻ 400 = 220²/R'

➻ 400 × R' = 48400

➻ R' = 48400/400

➻ R' = 121Ω

Now, calculating the total resistance of these two bulb as they are connected in parallel.

• 1/Req = 1/R + 1/R'

substitute the value we get

➻ 1/Req = 1/242 + 1/121

➻ 1/Req = 1+2/242

➻ 1/Req = 3/242

➻ Req = 242/3

➻ Req = 242/3 Ω

Now, calculating the current drawn by these bulb .

Using ohm's law

• V = IR

substitute the value we get

➻ 220 = I × 242/3

➻ I = 220×3/242

➻ I = 2.72 A

• Hence, the current drawn by these bulb will be 2.72 ampere.

Answer 2

Given :-

• Power of 1st bulb (P_{1}) = 200 W
• Power of 2nd bulb (P_{2}) = 400 W
• Potential difference (V) = 220 V

To Find :-

• Total current drawn by bulbs = ?

Solution :-

We know that,

✇ P = V²/R ✇ $\:$

Where,

• P is power
• V is potential difference
• R is Resistance

❏ Finding resistance of 1st bulb :-

↦ 200 = (220)²/R_{1}

↦ 200 × R_{1} = 48400

↦ R_{1} = 48400/200

➦ R_{1} = 242 Ω

❏ Finding resistance of 2nd bulb :-

⇒ 400 = (220)²/R_{2}

⇒ 400 × R_{2} = 48400

⇒ R_{2} = 48400/400

➠ R_{2} = 121 Ω

Now,

We know that,

✇ 1/Req = 1/R_{1} + 1/R_{2} ✇

❏ Finding equivalent resistance :-

↪ 1/Req = 1/242 + 1/121

↪ 1/Req = (1 + 2)/242

↪ 1/Req = 3/242

➢ Req = 242/3 Ω

Again we know that,

✇ Ohm's law : V = IR ✇

❏ Finding current drawn by bulbs :-

➝ 220 = I × 242/3

➝ I = 220 × 3/242

➡ I = 2.72 A

∴ Hence, the total current drawn by bulbs is 2.72 A.

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