An inductance coil has a reactance of 100 $\Omega$. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45o. The self-inductance of the coil is
Solution
L-R circuit :
<br><br>tan 45<sup>o</sup> = ${{{X_L}} \over R}$
<br><br>$\Rightarrow$ 1 = ${{{X_L}} \over R}$
<br><br>$\Rightarrow$ X<sub>L</sub> = R
<br><br>Now Z = $\sqrt {{R^2} + X_L^2}$
<br><br>or Z = $\sqrt {X_L^2 + X_L^2} = \sqrt {2X_L^2}$ = $\sqrt 2 {X_L}$
<br><br>$\Rightarrow$ 100 = $\sqrt 2 {X_L}$
<br><br>X<sub>L</sub> = ${{100} \over {\sqrt 2 }}$
<br><br>$\Rightarrow$ $\omega L$ = ${{100} \over {\sqrt 2 }}$
<br><br>$\Rightarrow$ L = ${{100} \over {\sqrt 2 \times 2 \times 3.14 \times 1000}}$
<br><br>= 1.1 $\times$ 10<sup>–2</sup> H
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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