Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240V, 10kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Exercise
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Here given,
Inductance,$ L = 0.50H$
Resistance, $\text{R}=100\Omega$
Rms value of voltage, $V_{rms} = 240V$
Frequency of Ac supply, $f = 10kHz = 10^4Hz$
Therefore,
Angular frequency, $\omega=2\pi\text{f}=2\pi\times10^4\text{rad }\text{s}^{-1}\ \text{Peak voltage},$
$\text{V}_0=\sqrt{2}\text{V}_{\text{rms}}$
$=\sqrt{2}\times240$
$= 339.36V$
Maximum current,
$\text{I}_0=\frac{\text{V}_0}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$
$=\frac{339.36}{\sqrt{(100)^2+(2\pi\times10^4\times0.5)^2}}\text{A}$
$=\frac{339.36}{31416}\text{A}$
$= 0.01212\ A = 1.12 \times 10^{-2}A$
This current is much smaller than for the low frequency case (1.82 A in above question), showing that the inductive reactance is very large at high frequencies and inductor in circuit nearly amounts to an open circuit.
In d.c. circuit (after steady state) $\omega=0.$
$\therefore\ \text{Z}_{\text{L}}=\omega\text{L}=0$
i.e., inductance L behaves like a pure inductor.
art

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