  Question

Two bulbs consume same energy when operated at 200 V and 300 V, respectively. When these bulbs are connected in series across a dc source of 500 V, then

A
ratio of potential difference across them is 3/2  B
ratio of potential difference across them is 4/9  C
ratio of power produced in them is 4/9  D
ratio of power produced in them is 2/3  Solution

The correct options are B ratio of potential difference across them is 4/9 C ratio of power produced in them is 4/9$$\displaystyle\,P\,=\,\frac{V^{2}}{R}$$  or  $$\displaystyle\,R\,=\,\frac{V^{2}}{P}$$  or  $$R\,\propto\,V^{2}$$, i.e., $$\displaystyle\, \frac{R_{1}}{R_{2}}\,=\,\left(\frac{200}{300}\right)^{2}\,=\,\frac{4}{9}$$.When connected in series, potential drop is in the ratio of their resistances. So, $$\displaystyle\,\frac{V_{1}}{V_{2}} \,=\,\frac{R_{1}}{R_{2}}\,=\,\frac{4}{9}$$ Now, $$P\,=\,I^{2}\,R$$    Or $$P\,\propto\,R$$ (in series I is the same) Or $$\displaystyle\frac{P_{1}}{P_{2}}\,=\,\frac{R_{1}}{R_{2}}\,=\,\frac{4}{9}$$   Physics

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