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Two bulbs one rated $100 W$ at $220 V$ and the other $200 W$ at $220 V$ , are connected in series to the electric supply at $220 V$ the current drawn by the two bulbs is [[1]] $A$ .

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Detailed Solution

When two bulbs of ratings

100 W

220 V

and

200 W

220 V

are connected in series to the electric supply at

220 V

then the current drawn by the two bulbs is

0.3 A

.
It is given that two bulbs of ratings

60 W

220 V

and

40 W

220 V

are connected in parallel combination with the mains at

220 V

.
Let the bulb with Power,

(P) = 100 W

and potential difference,

V = 220 V

be Bulb

1

.
Let the bulb with Power,
(

P) = 200 W

and potential difference,

V = 220 V

be Bulb

2

.
It is known that the power

(P)

of the component is given by the relation

P = \frac{V ²}{R}

R = \frac{V ²}{P}

Where

V

and

R

are the potential difference and the resistance respectively.
Therefore, the resistance of bulb

1

is given as

R_{1} = \frac{220^{2}}{100}

\Rightarrow R_{1} = 484 Ω

Similarly, the resistance of bulb

2

R_{2} = \frac{220^{2}}{200}

\Rightarrow R_{2} = 242 Ω

Since, they are connected in series, the equivalent resistance is equal to

R = R_{1} + R_{2}

\Rightarrow R = 484 + 242

\Rightarrow R = 726 Ω

Now, from the Ohm’s law, the current drawn is

I = \frac{V}{R}

On substituting the given value, we get the current as

\Rightarrow I = \frac{220}{726}

\Rightarrow I = 0.30 A

The total current drawn by the two bulbs is

0.30 A

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