• # question_answer Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in series. The powers dissipated in them have the ratio                                                                [NCERT 1977] A)            1 : 2                                           B)            2 : 1 C)            1 : 1                                           D)            1 : 4

In series, current is same in both the bulbs, hence $P\propto R\,(P={{i}^{2}}R)$  $\therefore \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}$