Obtain the answers to (a) and (b) in Exercise 13, if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence explain statement that at very high frequency, an inductor in circuit nearly amounts to an open circuit. How does an inductor behave in a d.c. circuit after the steady state?

Solution

Here given,
Inductance, L = 0.50 H
Resistance, R = 100

Ω

Rms value of voltage, V_{rms} = 240 V Frequency of Ac supply, f = 10 kHz = 10^{4} Hz

Therefore,

Angular frequency,

ω = 2 πf = 2 π \times 10^{4} rad s^{- 1}

Peak voltage,

V_{0} = \sqrt{2} V_{rms} = \sqrt{2} \times 240 = 339.36 V

Maximum current,

I_{0} = \frac{V_{0}}{\sqrt{R^{2} + ω^{2} L^{2}}}

= \frac{339.36}{\sqrt{(100)^{2} + {(2 π \times 10^{4} \times 0.5)}^{2}}} A = \frac{339.36}{31416} A (Neglecting R) = 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) ω = 0.
∴ Z_L = ωL = 0
i.e., inductance L behaves like a pure inductor.

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