- Home
- Standard 12
- Physics
7.Alternating Current
hard
કોઈલનો ઈમ્પિડન્સ $100\, \Omega$ છે. $1000\, Hz$ ની આવૃતિ કોઈલ પર લગાવતા વૉલ્ટેજ પ્રવાહ કરતા $45^{\circ}$ આગળ છે. તો કોઇલનો ઇન્ડકટન્સ
A
$1.1 \times 10^{-2}\; H$
B
$1.1 \times 10^{-1} \;H$
C
$5.5 \times 10^{-5} \;H$
D
$6.7 \times 10^{-7}\; H$
(JEE MAIN-2020)
Solution
Reactance of inductance coil
$=\sqrt{{ R ^{2}+ x _{ L }^{2}}}=100$ $…(i)$
$f =1000 Hz$ of applied $AC$ signal
Voltage leads current bly $45^{\circ}$
$\tan 45^{\circ}=\frac{ i X _{ L }}{ iR }=\frac{\omega L }{ R }$
ie $R = X _{ L }=\omega L$
Putting in eqn $(i):$ $\sqrt{ X _{ L }^{2}+ X _{ L }^{2}}=100$
$\sqrt{2} X _{ L }=100 \Rightarrow X _{ L }=50 \sqrt{2}$
ie $\omega L =50 \sqrt{2}$
$L =\frac{50 \sqrt{2}}{\omega}=\frac{50 \sqrt{2}}{2 \pi f }=\frac{25 \sqrt{2}}{\pi \times 1000} H$
$=1.125 \times 10^{-2} H$
Standard 12
Physics
