Question

Two bulbs of 100watt & 200 watt are connected to 220 volt battery separately , after that

both the bulbs are connected in series with a 200 volt battery . find total current flowing in a

circuit when both are connected in series .​

Answer 1

Answer :-

Total current flowing in the circuit when both bulbs are connected in series is 0.275 A.

Explanation :-

We have :-

→ Power of the 1st bulb = 100 W

→ Power of the 2nd bulb = 200 W

→ 1st Potential difference = 220 V

→ 2nd Potential difference = 200 V

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Resistance of the 1st bulb :-

⇒ P₁ = V²/R₁

⇒ R₁ = V²/P₁

⇒ R₁ = (220)²/100

⇒ R₁ = 484 Ω

Resistance of the 2nd bulb :-

⇒ P₂ = V²/R₂

⇒ R₂ = V²/P₂

⇒ R₂ = (220)²/200

⇒ R₂ = 242 Ω

Equivalent resistance for series connection :-

⇒ R = R₁ + R₂

⇒ R = (484 + 242) Ω

⇒ R = 726 Ω

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Now, we will calculate the current flowing in the circuit for series connection ny using 'Ohm's Law'.

V = IR

⇒ I = V/R

⇒ I = 200/726

⇒ I = 0.275 A

Answer 2

Given :-

Two bulbs of 100watt & 200 watt are connected to 220 volt battery separately , after that

both the bulbs are connected in series with a 200 volt battery .

To Find :-

Current

Solution :-

We know that

R = V²/P

R = (220)²/100

R = 48400/100

R = 484 \Omega

Now

R = V²/P

R = (220)²/200

R = 48400/200

R = 484/2

R = 242 \Omega

Now

I = V/R

I = 200/(484 + 242)

I = 200/726

I = 0.275 ampere