JEE MAINS PHYSICS QUESTION SEPTEMBER 2020
Attachments:
![](https://hi-static.z-dn.net/files/d67/b43615f6cd7e5799820590000e0aed75.jpg)
Answers
Answered by
0
The circuit on at t = 0, find the time when energy speed in inductor becomes 1/n times of maximum energy stored in it.
solution : energy stored in inductor is given by, U = 1/2 Li²
here inductance L remains constant
so, U ∝ I²
i.e., U₁/U₂ = (I₁/I₂)²
here let U₂ = U₀ = maximum energy stored in inductor
so, U₁/U₂ = 1/n
we know, I = I₀[1 - e¯Rτ/L ]
so, I₁ = I₂ [1 - e¯Rτ/L] [ as I₂ = I₀]
⇒I₁/I₂ = (1 - e¯Rτ/L)
so, 1/n = (1 - e¯Rτ/L)²
⇒1/√n = (1 - e¯Rτ/L)
⇒(√n - 1)/√n = e¯Rτ/L
⇒ln[(√n - 1)/√n] = -Rτ/L
⇒Lln[√n/(√n - 1)] = Rτ
but τ = (L/R) ln [√n/(√n - 1) ]
Therefore time when energy speed in inductor becomes 1/n times of maximum energy stored in it is (L/R) ln [√n/(√n - 1) ]
Similar questions
English,
5 months ago
History,
5 months ago
Computer Science,
1 year ago
Math,
1 year ago
Math,
1 year ago