JEE Advanced 2018 Paper 1 · Physics · Current Electricity · hard

A battery of emf 10 V and internal resistance 0.5 Ω is connected to an infinitely long ladder network. Each section consists of a 4 Ω resistor in the…

Numerical answer

A battery of emf

10\ \mathrm{V}

and internal resistance

0.5\ \Omega

is connected to an infinitely long ladder network. Each section consists of a

4\ \Omega

resistor in the series arm followed by a

4\ \Omega

resistor connected across the two rails. Find the power dissipated, in watts, in the first shunt (

4\ \Omega

across-the-rails) resistor, the one nearest the battery. Give the answer correct to two decimal places.

Show answer and solution

Answer: 3.14

Let

R_\infty

be the resistance of the infinite ladder. Removing the first section leaves an identical ladder, so

R_\infty=4+\dfrac{4R_\infty}{4+R_\infty}

, which gives

R_\infty^2-4R_\infty-16=0

and

R_\infty=2(1+\sqrt5)=6.472\ \Omega

. The current from the battery is

I=\dfrac{10}{6.472+0.5}=1.4343\ \mathrm{A}

. The first series resistor drops

4I=5.737\ \mathrm{V}

(from the series arm), so the potential across the first shunt resistor, which is also across the rest of the ladder, is

V=I(R_\infty-4)=1.4343\times2.472=3.546\ \mathrm{V}

. Power in the shunt resistor is

P=V^2/4=3.14\ \mathrm{W}

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