Question

A solonoid of 10 henrey and 2ohm resistance is connected to a10volt battary .in how much time the magnetic energy will increase from 0to1/4of maximum energy

Answer 1

SoluTion :-

Given -

L = 10 H

R = 2 ohm

V = 10 volt

Now, the maximum current is

$\sf {i_o =\frac{V}{R} }\\\\\sf {i_o =\frac{10}{2} }\\\\\sf {i_o =5A}$

The maximum energy is

$\sf {E_o=\frac{1}{2}Li\overset{\sf{2}}{\sf{0}} }\\\\\sf {E_o=\frac{1}{2} \times 10 \times 5 \times 5}\\\\\sf {E_o=125\ J}$

Now, the magnetic energy

$\sf {E=\frac{E_o}{4} }\\\\\sf {E=\frac{125}{4}J}$

Now,

$\sf {E=\frac{1}{2}Li^2 }\\\\\sf {\frac{125}{4}= \frac{1}{2}Li^2}\\\\\sf {i^2 = \frac{125}{2 \times 10} }\\\\\sf {i^2 = 2.5\ A}$

Time taken to rise current from 0 - 2.5 A.

The instantaneous current during its growth in an L-R circuit.

$\sf {i=i_o(1-e^-{\frac{Rt}{L} })}\\\\\\\sf {2.5=5(1-e^-{\frac{Rt}{L} })}\\\\\\\sf {e^-{\frac{Rt}{L}=0.5}$

$\sf {-{\frac{Rt}{L}=In(0.5)}$

$\sf {\frac{Rt}{L}=0.693}$

$\sf {t=\frac{6.93}{2} }$

$\sf {t=3.5\ seconds}$