KVPY SA Exam 2015 Question Paper

Question : 26 of 80

Marks: +1, -0

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 drives current in the circuit. What should be the value of r such that the bulb dissipates about 1 W of power.

14.8Ω
29.6Ω
7.4Ω
3.7Ω

Validate

Solution:

Let assume

R_\eq = PQ = x

R_\eqPQ=r+{rx}/{r+x}

x={r^2+rx+rx}/{r+x}

rx + x^2 = r ^2 + 2rx

x ^2 - rx - r^ 2 = 0

x={+r±√{r^2+4r^2}}/2⇒{r(1+√5)}/2

Power in bulb = 1 watt

i^ 2R = 1

i^ 2 × 16 = 1

i-1/4

amp.

i=10/{R+R_\PQ}

1/4=10/{16+r/2(1+√5)

16+r/2(1+√5)=40

r = 14.8 Ω

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