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Question
Obtain if the circuit is connected to a high-frequency supply (240 V, 10 kHz). 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?
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Solution
Inductance of the inductor, L = 0.5 Hz
Resistance of the resistor, R = 100 Ω
Potential of the supply voltages, V = 240 V
Frequency of the supply, v = 10 kHz = 104 Hz
Angular frequency, ω = 2πv = 2π × 104 rad/s
(a) Peak voltage, V0 = `sqrt2 xx "V" = 240 sqrt2 "V"`
Maximum current, I0 = `"V"_0/sqrt("R"^2 + ω^2"L"^2)`
= `(240 sqrt2)/sqrt((100)^2 + (2π xx 10^4)^2 xx (0.50)^2)`
= 1.1 × 10−2 A
(b) For phase difference Φ, we have the relation:
`tan phi = (ω"L")/"R"`
= `(2π xx 10^4 xx 0.5)/100`
= 100π
`phi` = 89.82° = `(89.82π)/180 "rad"`
ωt = `(89.82π)/180`
t = `(89.82π)/(180 xx 2π xx 10^4)`
= 25 μs
It can be observed that I0 is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.
In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.
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