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7.Alternating Current
normal
Consider the infinite ladder circuit shown below.For which angular frequency $\omega$ will the circuit behave like a pure inductance?
A$\frac{L C}{\sqrt{2}}$
B$\frac{1}{\sqrt{L C}}$
C$\frac{2}{\sqrt{L C}}$
D$\frac{2 L}{\sqrt{C}}$
(KVPY-2010)
Solution
(d)In an infinite network, adding one more loop does not affects total impedence of circuit.
$\Rightarrow \quad Z_{C D}=Z_{A B}$
$\Rightarrow(Z \| C)$ series $L=Z$
$\Rightarrow \frac{Z X_C}{Z+X_C}+X_L=Z$
$\Rightarrow \quad \frac{Z / \omega C}{Z+\frac{1}{\omega C}}=Z-\omega L$
$\Rightarrow \quad \frac{\frac{Z}{\omega C}}{Z+\frac{1}{\omega C}}=Z-\omega L$
$\Rightarrow \quad Z^2-\omega L Z-\frac{L}{C}=0$
$\Rightarrow$ $Z=\omega L \pm \sqrt{\omega^2 L^2+\frac{4 L}{C}}$
Now, $Z=\omega L$ or circuit is purely inductive when,
$\omega^2 L^2+\frac{4 L}{C}=0 \Rightarrow \omega^2=-\frac{4}{L C}$
But this is not possible or $\omega$ is imaginary.
Hence, circuit does not behaves like an inductor.
Standard 12
Physics
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