Question 140: AC fundamentals

Question

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 45
o
. The self-inductance of the
coil is

Accepted Answer

L-R circuit :
tan 45
o
= ${{{X_L}} \over R}$
$ \Rightarrow $ 1 = ${{{X_L}} \over R}$
$ \Rightarrow $ X
L
= R
Now Z = $\sqrt {{R^2} + X_L^2} $
or Z = $\sqrt {X_L^2 + X_L^2}  = \sqrt {2X_L^2} $ = $\sqrt 2 {X_L}$
$ \Rightarrow $ 100 = $\sqrt 2 {X_L}$
X
L
= ${{100} \over {\sqrt 2 }}$
$ \Rightarrow $ $\omega L$ = ${{100} \over {\sqrt 2 }}$
$ \Rightarrow $ L = ${{100} \over {\sqrt 2  	imes 2 	imes 3.14 	imes 1000}}$
= 1.1 $ 	imes $ 10
Ã¢ÂÂ2
H

Answer

6.7 $ 	imes $ 10
–7
H

Answer

1.1 $ 	imes $ 10
–1
H

Answer

5.5 $ 	imes $ 10
–5
H

Answer

1.1 $ 	imes $ 10
–2
H

AC fundamentals

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 45 o . The self-inductance of the coil is

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