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Question

A light bulb of resistance r = 16Ω is attached in series with an infinite resistor network with identical resistances r as shown below. A 10 V battery derives current in the circuit. What should be the value of r such that the bulb dissipated about 1 W of power?


A

14.8Ω

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B

29.6Ω

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C

7.4Ω

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D

3.7Ω

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Solution

The correct option is A

14.8Ω


Pbulb=v2R=i2R

I=v216

VB= 4 V

I =i2×16

IB=14 Amp.

6=14×req (equivalent of groups of r)

where

req=r+req.rreq+r


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