In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. Ea2bJ/s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of ba will be _____.

E=12LI2Rate of storing energy=dE/dt=LIdldtNow for L-R circuitI=ER1−e−tRL So dIdt=ELe−tRLdEdt=E2R1−e−tRLe−tRLTime at which it will be maximumt=LRln2substituting in above equation So dEdt=E2R1−12×12⇒E24R=E2100=E22×50a=2,b=50 So ba=25

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. $\frac{E^{a}}{2 b} J / s$ represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of $\frac{b}{a}$ will be _____.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 25.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$E = \frac{1}{2} {LI}^{2}$

$Rate of storing energy$

$= dE / dt = LI (\frac{dl}{d t})$

Now for L-R circuit

$I = \frac{E}{R} (1 - e^{- t \frac{R}{L}})$

$So \frac{d I}{d t} = \frac{E}{L} e^{- \frac{t R}{L}}$

$\frac{d E}{d t} = \frac{E^{2}}{R} (1 - e^{- \frac{t R}{L}}) (e^{- t \frac{R}{L}})$

Time at which it will be maximum

$t = \frac{L}{R} \ln (2)$

substituting in above equation

$So \frac{d E}{d t} = \frac{E^{2}}{R} (1 - \frac{1}{2}) \times \frac{1}{2}$

$\Rightarrow \frac{E^{2}}{4 R} = \frac{E^{2}}{100} = \frac{E^{2}}{2 \times 50}$

$\begin{aligned} a = 2, b = 50 \\ So \frac{b}{a} = 25 \end{aligned}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th