Question

A solenoid of 10 Henry inductance and 2 ohm resistance, is connected to a 10 volt battery. In how much time the magnetic energy will be reaches to 1/4th of the maximum value?

• A. 3.5 sec
• B. 2.5 sec
• C. 5.5 sec
• D. 7.5 sec

Answer 1

The correct option is A 3.5 sec

Given that,

L = 10 H

R = 2 ohm

V = 10 volt

Now, the maximum current is

$i_0=\frac{V}{R}$

$i_0=\frac{10}{2}$

$i_0=5A$

The maximum energy is

$E_0=\frac{1}{2}L{i}_0^2$

$E_0=\frac{1}{2}\times 10\times 5\times 5$

$E_0=125J$

Now, the magnetic energy

$E=\frac{E_0}{4}$

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

Now,

$E=\frac{1}{2}L{i}^2$

$\frac{125}{4}=\frac{1}{2}L{i}^2$

$\frac{125}{2\times 10}=i^2$

$i^2=\frac{25}{2}$

$i=\frac{5}{2}$

$i=2.5A$

Now, the time taken to rise current from 0 - 2.5A.

We know that, the instantaneous current during its growth in an L-R circuit.

$i=i_0(1-e^{-\frac{Rt}{L}})$

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

$e^{\frac{-Rt}{L}}=0.5$

$\frac{-Rt}{L}=\ln \left(0.5\right)$

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

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

$t=3.46$

$t=3.5\sec$

Hence, the magnetic energy will be increases to $\frac{1}{4}$ of the maximum value at $3.5$ sec.