Two bulbs 25 W, 100 V (upper bulb in figure) and 100 W, 200 V (lower bulb in figure) are connected in the circuit as shown in figure. Choose the correct answer(s).

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Ratio of heat produced per second in bulb will be 1:1

Ratio of heat produced in branch AB to that produced in branch CD will be 1:2

Current drawn from the cell is 0.4 A

Heat lost per second in the circuit will be 80 J

answer is A, D.

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Detailed Solution

Resistance of upper bulb, $R_{1} = 100^{2} / 25 = 400 Ω$

Resistance of lower bulb, $R_{2} = 200^{2} / 100 = 400 Ω$

Equivalent resistance of the circuit, $R_{e q} = 500 Ω$

Heat lost per second in the circuit is

$e^{2} / R_{e q} = 200^{2} / 500 = 80 J s^{- 1}$

Potential difference across branches AB and CD will be same. Hence, ratio of heat generated in them is

$\frac{H e a t_{A B}}{H e a t_{C D}} = \frac{R_{C D}}{R_{A B}} = \frac{1000}{500} = 2 : 1$

Hence, 3 is incorrect.

Since potential difference across branches AB and CD will be the same, but their resistances are different, so current in them will be different. As the resistances of the bulbs are same, so heat generated in them will be different. Hence, 2 is incorrect. Current drawn from the cell is $e / R_{e q} = 200 / 500 = 0.4 A$

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