Consider the infinite ladder circuit shown be | vedclass

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(C) For an infinite ladder network,the total impedance $Z$ remains unchanged if one more section is added to the input.Let $Z$ be the equivalent imped

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$\frac{2}{\sqrt{L C}}$

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(C) For an infinite ladder network,the total impedance $Z$ remains unchanged if one more section is added to the input.Let $Z$ be the equivalent impedance of the infinite ladder. The circuit consists of an inductor $L$ in series with the parallel combination of a capacitor $C$ and the rest of the infinite ladder $Z$.Thus,$Z = j\omega L + \frac{Z \cdot (1/j\omega C)}{Z + (1/j\omega C)}$.$Z = j\omega L + \frac{Z}{1 + j\omega C Z}$.$Z(1 + j\omega C Z) = j\omega L(1 + j\omega C Z) + Z$.$Z + j\omega C Z^2 = j\omega L - \omega^2 L C Z + Z$.$j\omega C Z^2 + \omega^2 L C Z - j\omega L = 0$.Dividing by $j\omega C$,we get $Z^2 + \frac{\omega L}{j} Z - \frac{L}{C} = 0$,which is $Z^2 - j\omega L Z - \frac{L}{C} = 0$.Solving for $Z$ using the quadratic formula: $Z = \frac{j\omega L \pm \sqrt{(j\omega L)^2 - 4(1)(-L/C)}}{2} = \frac{j\omega L \pm \sqrt{-\omega^2 L^2 + 4L/C}}{2}$.For the circuit to behave as a pure inductor,the impedance $Z$ must be purely imaginary,i.e.,$Z = j\omega L_{eq}$.This requires the term under the square root to be positive,and the real part to be zero. However,the expression shows that for the circuit to have a real part of zero,the condition depends on the frequency range. Specifically,the cut-off frequency is $\omega_c = \frac{2}{\sqrt{LC}}$. Below this frequency,the circuit has a real component (resistance/attenuation). Above this frequency,it behaves as a pure reactance. The question asks for the condition of pure inductance,which is satisfied in the pass-band where the characteristic impedance is purely imaginary. The correct value for the cut-off frequency is $\omega = \frac{2}{\sqrt{LC}}$.

Consider the infinite ladder circuit shown below. For which angular frequency $\omega$ will the circuit behave like a pure inductance?

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