JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 1)
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
6.7 $$ \times $$ 10–7 H
1.1 $$ \times $$ 10–1 H
5.5 $$ \times $$ 10–5 H
1.1 $$ \times $$ 10–2 H
Explanation
L-R circuit :
tan 45o = $${{{X_L}} \over R}$$
$$ \Rightarrow $$ 1 = $${{{X_L}} \over R}$$
$$ \Rightarrow $$ XL = 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 = $$\sqrt 2 {X_L}$$
XL = $${{100} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$\omega L$$ = $${{100} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ L = $${{100} \over {\sqrt 2 \times 2 \times 3.14 \times 1000}}$$
= 1.1 $$ \times $$ 10–2 H
tan 45o = $${{{X_L}} \over R}$$
$$ \Rightarrow $$ 1 = $${{{X_L}} \over R}$$
$$ \Rightarrow $$ XL = 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 = $$\sqrt 2 {X_L}$$
XL = $${{100} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$\omega L$$ = $${{100} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ L = $${{100} \over {\sqrt 2 \times 2 \times 3.14 \times 1000}}$$
= 1.1 $$ \times $$ 10–2 H
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