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  • 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.

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)

best_answer

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.

Posted by

Pankaj Sanodiya

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