Ncert Exercises & Solutions

7.14 Obtain the answers (a) to (b) in Exercise 7.13 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?

Given,

Inductance, 𝐿 = 0.50 H

Resistance, 𝑅 = 100 Ω

rms voltage, 𝑉rms = 240 V

frequency, 𝑓 = 10 kHz

(a) The maximum current in the coil,

$I_{m} = \frac{\sqrt{2} \times V_{r m s}}{\sqrt{X_{L}^{2} + R^{2}}}$

For an inductive circuit

$X_{L} = 2 π \times 10 k H z \times 0.5 H = 31415.9$

So, The maximum current in the coil,

$I_{m} = \frac{\sqrt{2} \times 240 V}{\sqrt{(31415.9 Ω)^{2} + (100 Ω)^{2}}} = 1.1 \times 10^{- 2} A$

(b)

$\begin{array}{l} \tan ϕ = \frac{ω L}{R} = \frac{2 π \times 10^{4} \times 0.5}{100} = 100 π \\ ϕ = \tan^{- 1} 100 π = {89.82}^{\circ} = \frac{89.82 π}{180} \\ Timelag = \frac{ϕ}{ω} = \frac{89.82 π}{180 \times 2 π \times 50} = 25 μ S \end{array}$

𝐼m in this case, is too small, so it can be concluded that at high frequencies an
inductor behaves as the open circuit.

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