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)

Answer :

We are given with the ratio of resistance of the two bulbs, let it be R_{1}: R_{2} = 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 bulb_{1} be E_{1} and bulb_{2} be E_{2}.

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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