3.21 Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

    

Answers (1)

First, let us find the equivalent of the infinite network,

let equivalent resistance = R'

Here from the figure, We can consider  the box as a resistance of R'

Now, we can write,

Equivalent resistance = R'

' =[( R')Parallel with (1)] + 1 + 1

  \frac{R'*1}{R'+1}+2=R'

   R'+2R'+2=R'^2+R'

R'^2-2R'-2=0

R'=1+\sqrt{3},or1-\sqrt{3}

Since resistance can never be negative we accept

 R'=1+\sqrt{3}

, We have calculated the equivalent resistance of infinite network,

NOW

Total Equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network

                                            = 0.5+1+1.73

                                            =3.23 ohm

V=IR

I=\frac{V}{R}=\frac{12}{3.23}=3.72A

Hence current drawn from the 12V battery is 3.72 Ampere.

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