Q. 2

# Two electric bulbs have resistances in the ratio 1 : 2. If they are joined in series, the energy consumed in these are in the ratio _________. (1 : 2, 2 : 1, 4 : 1, 1 : 1)

We are given with the ratio of resistance of the two bulbs, let it be R1: R2 = 1 : 2 , the bulbs are connected in series , so same amount of current will pass through both of the bulbs. The energy consumed by a bulb is given by,

E(electric energy) = V(potential difference) X I(current) X T(time)

As we do not know about the potential difference in this question but we know about the current (same in both bulbs) , so in order to solve this question we need to change the equation in the form of current , resistance and time only.

Put V = IR form Ohm’s law relation in this electric energy equation.

E = (IR) × I × T = , which is similar form of Joule’s law of heating.

Let the electric energies of bulb1 be E1 and bulb2 be E2.

As time taken by the current to pass through both bulbs is always same , The ratio of the electric energies is same as the ratio of resistances of the corresponding bulbs.

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