JEE Main 2020 — Alternating Current Question with Solution
From: JEE Main 2020 (Online) 2nd September Evening Slot
Question
An inductance coil has a reactance of 100 .
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
Choose an option
Show full solutionCorrect option: D
Correct answer
D1.1 10–2 H
Step-by-step explanation
L-R circuit :
tan 45o =
1 =
XL = R
Now Z =
or Z = =
100 =
XL =
=
L =
= 1.1 10–2 H
tan 45o =
1 =
XL = R
Now Z =
or Z = =
100 =
XL =
=
L =
= 1.1 10–2 H
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