Determine the current (in A) drawn from a 12; | vedclass

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(C) Let the equivalent resistance of the infinite network be $R^{\prime}$.Since the network is infinite,adding one more section of the same structure

Vedclass · Accepted Answer

$3.72$

Vedclass · Answer

(C) Let the equivalent resistance of the infinite network be $R^{\prime}$.Since the network is infinite,adding one more section of the same structure will not change the equivalent resistance.The network consists of two $1\; \Omega$ resistors in series with the rest of the infinite network,which is in parallel with a $1\; \Omega$ shunt resistor.Thus,$R^{\prime} = 1 + 1 + \frac{R^{\prime} 	imes 1}{R^{\prime} + 1} = 2 + \frac{R^{\prime}}{R^{\prime} + 1}$.Multiplying by $(R^{\prime} + 1)$,we get $R^{\prime}(R^{\prime} + 1) = 2(R^{\prime} + 1) + R^{\prime}$.$(R^{\prime})^2 + R^{\prime} = 2R^{\prime} + 2 + R^{\prime}$.$(R^{\prime})^2 - 2R^{\prime} - 2 = 0$.Using the quadratic formula $R^{\prime} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $R^{\prime} = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.Since resistance cannot be negative,$R^{\prime} = 1 + \sqrt{3} \approx 1 + 1.732 = 2.732\; \Omega$.The total resistance of the circuit including the internal resistance $r = 0.5\; \Omega$ is $R_{total} = R^{\prime} + r = 2.732 + 0.5 = 3.232\; \Omega$.The current drawn from the $12\; V$ supply is $I = \frac{V}{R_{total}} = \frac{12}{3.232} \approx 3.713\; A$.Rounding to two decimal places,the current is $3.72\; A$.

Determine the current (in $A$) drawn from a $12\; V$ supply with internal resistance $0.5\; \Omega$ by the infinite network shown in the figure. Each resistor has $1\; \Omega$ resistance.

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