JEE PYQ: Alternating Current Question 20
Question 20 - 2020 (02 Sep Shift 2)
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 $45°$. The self-inductance of the coil is:
(a) $1.1 \times 10^{-2}$ H (b) $1.1 \times 10^{-1}$ H (c) $5.5 \times 10^{-5}$ H (d) $6.7 \times 10^{-7}$ H
Show Answer
Answer: (a)
Solution
$\tan 45° = 1 \Rightarrow X_L = R$. $Z = \sqrt{2}R = 100 \Rightarrow R = 50\sqrt{2}$. $L = \frac{50\sqrt{2}}{2\pi \times 1000} = 1.1 \times 10^{-2}$ H.
BETA
