Answer :

**Concepts/Formulas Used:**

__Resistors in Series:__

__Power dissipated by a resistor:__

The power dissipated by a resistor of resistance R with potential difference V across it and current I passing through it is given by:

__Ohm’s law:__

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Let us find the resistance of the two bulbs.

We know that,

Hence,

(i)

In series, the net resistance is:

The current in the circuit is

The power dissipated in the first bulb:

The power dissipated in the first bulb:

(ii)

In parallel, the potential through both is V. Hence the power is the same as the rated power i.e. P_{1} and P_{2}.

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