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Question

A torch bulb rated as 4.5 W1.5 V is connected as shown in the figure. The emf of the cell needed to make the bulb glow at full intensity is

A
4.5 V
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B
1.5 V
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C
2.67 V
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D
13.5 V
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Solution

The correct option is D 13.5 V

Given, P=4.5 W,V=1.5 V
current passing through the bulb= 4.51.5=3 A

In the given situation, potential difference across bulb is due to potential difference due to 1 Ω resistor because of parallel connection.
Hence, current in 1 Ω resistor=1.51=1.5 A

itotal=1.5+3=3.5A
EMF of the battery =V=eitotal×rinternale=1.5+4.5×2.67=13.5 V

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