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?

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Payal Gupta | Contributor-Level 10

8 months ago

7.14 Inductance of the Inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of supply voltage, V = 240 V

Frequency of the supply, $ν$ = 10 kHz = 10 $* 10^{3}$ Hz

Peak voltage is given as $V_{0}$ = $\sqrt[]{2 V} = \sqrt[]{2} * 240$ = 339.41 V

Angular frequency of the supply, $ω = 2 π ν$ = 2 $π * 10^{4} r a d / s$

Maximum current in the supply is given as

$I_{0}$ = $\frac{V_{0}}{\sqrt[]{R^{2} + {(ω L)}^{2}}}$ = $\frac{339.41}{\sqrt[]{100^{2} + {(2 π * 10^{4} * 0.50)}^{2}}}$ = 10.80 $* 10^{- 3}$ A

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Physics Ncert Solutions Class 12th 2023

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