Search any question & find its solution
Question:
Answered & Verified by Expert
The time constant of inductance coil is $3 \mathrm{~m} \mathrm{~s}$. When a $90 \Omega$ resistance is joined in series, then the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$. The inductance and the resistance of the coil are
Options:
Solution:
1669 Upvotes
Verified Answer
The correct answer is:
$54 \mathrm{mH}, 18 \Omega$
Time constant $t=\frac{L}{R}$
$3=\frac{L}{R}$ ...(i)
When a $90 \Omega$ resistance is joined in series the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$.
So, $0.5=\frac{L}{R+90}$ ...(ii)
From Eqs. (i) and (ii)
$\frac{3 \times 10^{-3}}{0.5 \times 10^{-3}}=\frac{R+90}{R}$
$\frac{30}{5}=\frac{R+90}{R}$
$6 R=R+90$
$R=18 \Omega$
This value put in Eq. (i)
$3 \times 10^{-3}=\frac{L}{R}$
$3 \times 10^{-3}=\frac{L}{18}$
or $\quad L=54 \mathrm{mH}$
$3=\frac{L}{R}$ ...(i)
When a $90 \Omega$ resistance is joined in series the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$.
So, $0.5=\frac{L}{R+90}$ ...(ii)
From Eqs. (i) and (ii)
$\frac{3 \times 10^{-3}}{0.5 \times 10^{-3}}=\frac{R+90}{R}$
$\frac{30}{5}=\frac{R+90}{R}$
$6 R=R+90$
$R=18 \Omega$
This value put in Eq. (i)
$3 \times 10^{-3}=\frac{L}{R}$
$3 \times 10^{-3}=\frac{L}{18}$
or $\quad L=54 \mathrm{mH}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.