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

ratio of potential difference across them is 3/2

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ratio of potential difference across them is 4/9

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ratio of power produced in them is 4/9

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ratio of power produced in them is 2/3

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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
$P = \frac{V^{2}}{R}$ or $R = \frac{V^{2}}{P}$ or $R \propto V^{2}$ ,
i.e., $\frac{R_{1}}{R_{2}} = {(\frac{200}{300})}^{2} = \frac{4}{9}$ .
When connected in series, potential drop is in the ratio of their resistances. So, $\frac{V_{1}}{V_{2}} = \frac{R_{1}}{R_{2}} = \frac{4}{9}$
Now, $P = I^{2} R$ Or $P \propto R$ (in series I is the same)

Or $\frac{P_{1}}{P_{2}} = \frac{R_{1}}{R_{2}} = \frac{4}{9}$

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Similar questions

Ω

Ω

V_{A}

V_{B}

Column - I	Column -II
a. In series connection, the ratio of potential difference across A and B	$\frac{R_{A}}{R_{B}}$
b. In series connection the ratio of power consumed by A and B	$\frac{V_{A}^{2}}{V_{B}^{2}}$
c. In parallel connection, the ratio of current flowing through A and B	$\frac{R_{B}}{R_{A}}$
d. In parallel connection, the ratio of power connsumed in A and B	$\frac{V_{B}^{2}}{V_{A}^{2}}$

25 W, 40 W, 60 W

100 W

List I	List II
(P) $25 W$	(1) Highest potential differences
(Q) $100 W$	(2) Second highest potential difference
(R) $60 W$	(3) Third highest potential difference
(S) $40 W$	(4) Lowest potential difference

110 V

50 W

100 W

110 V

50 W

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