Question 212: 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 45o. 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 45o. The self-inductance of the coil is

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