Question

A solenoid has an inductance of 10 henry and a resistance of 2 ohm. It is connected to a 10 volt battery. How long will it take for the magnetic energy to reach 1/4 of its maximum value?

• A. $2.142s$
• B. $3.465s$
• C. $0.693s$
• D. $4.345s$

Answer 1

Answer: (B)

$3.465s$

The current grows in L - R circuit till steady state is reached.
At steady state, energy stored in inductor is macimum.
At some instant of time, current = I
$\therefore Maximumenergy=\frac{1}{2}L{I}_0^2$
Energy stored when current is $I=\frac{1}{2}L{I}^2$
$\therefore \frac{U_{max}}{U}=\frac{2L{I}_0^2}{2L{I}^2}$
or $\frac{U_{max}}{\left(\frac{U_{max}}{4}\right)}={\left(\frac{I_0}{I}\right)}^2$
or ${\left(\frac{I_0}{I}\right)}^2=4$
or $I=\frac{I_0}{2}$
For growth of current in L - R circuit,
$I=I_0[1-e^{-\frac{R}{I}t}]or\frac{I_0}{2}=I_0[1-e^{-\frac{2}{to}t}]$
or $\frac{1}{2}=1-e^{-t/5}$
or $e^{-t/5}=\frac{1}{2}$
or $t=5In2$
or $t=5\times 0.693$
or $t=3.465s$.