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?

6 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 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 $\times 10^{3}$ Hz

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

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

Maximum current in the supply is given as

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

Phase angle is also given by the relation,

$\tan ? \emptyset$ = $\frac{ω L}{R}$ = $\frac{2 π \times 10^{4} \times 0.5}{100}$ = 314.16

$\emptyset = 89.82 °$

= $\frac{89.82 \times π}{180}$ rad = 1.568 rad

$ω t = 1.568$

t = $\frac{1.568}{ω}$ = $\frac{1.568}{2 π \times 10^{4}}$ = 24.95 $\times 10^{- 6}$ s = 24.95 $μ$ s

It can be observed that $I_{0}$ is very small in this case. Hence, at

...more