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An infinite ladder network is made by repeating a unit consisting of a 4 Ω series resistor followed by a 3 Ω resistor connected across the line. A 12…

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An infinite ladder network is made by repeating a unit consisting of a

4\ \Omega

series resistor followed by a

3\ \Omega

resistor connected across the line. A

12\ \mathrm{V}

battery is connected across the input terminals of the ladder. The power dissipated in the first

3\ \Omega

shunt resistor is

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Answer: (B) $\dfrac{16}{3}\ \mathrm{W}$

Let the input resistance be

x

. Removing one unit leaves the same infinite ladder, so

x=4+\dfrac{3x}{3+x}

, which gives

x^{2}-4x-12=0

and

x=6\ \Omega

. Input current

=12/6=2\ \mathrm{A}

. Voltage across the first shunt (in parallel with the rest of the ladder,

3\parallel6=2\ \Omega

) is

2\times2=4\ \mathrm{V}

, equivalently

12-2\times4=4\ \mathrm{V}

. Power in the

3\ \Omega

resistor

=\dfrac{4^{2}}{3}=\dfrac{16}{3}\ \mathrm{W}

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