91Ó°ÊÓ

i) The inductor of inductance is$L_1=5.00\mathrm{mH}=5.00×{10}^{-3}H$

ii) The resistor of resistance is$R_1=25.0\mathrm{Î\copyright }=20×{10}^{-3}\mathrm{Î\copyright }$

iii) The inductor of inductance is$L_2=3.00\mathrm{mH}=3.00×{10}^{-3}H$

iv) The resistor of resistance is $R_2=30.0\mathrm{Î\copyright }$

v) The ratio of the magnetic flux is$\frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}=1.50$

We can use the concept of the decay current and expression of the inductive time constant.

Formulae:

$$\begin{gathered} i=i_0{e}^{-\frac{t}{r_L}} \\ \mathrm{Ï•}={\mathrm{Ï•}}_0{e}^{-\frac{t}{r_L}} \\ t=\frac{L}{R} \end{gathered}$$

The time at which the flux through a turn in two inductors equal:

The expression of decay of current is

$i=i_0{e}^{-\frac{t}{r_L}}$

Then the flux will decay as,

$$\begin{gathered} {\mathrm{Ï•}}_1={\mathrm{Ï•}}_{01}{e}^{-\frac{t}{t_L}} \\ {\mathrm{Ï•}}_2={\mathrm{Ï•}}_{02}{e}^{-\frac{t}{t_L}} \end{gathered}$$

The expression of the inductive time constant is

$t=\frac{L}{R}$

According to the condition,

$$\begin{gathered} {\mathrm{Ï•}}_1={\mathrm{Ï•}}_2 \\ {\mathrm{Ï•}}_{01}{e}^{-\frac{t}{t_{L1}}}={\mathrm{Ï•}}_{02}{e}^{-\frac{t}{t_{L2}}} \\ \frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}=\frac{e^{-\frac{t}{t_{L1}}}}{e^{-\frac{t}{t_{L2}}}} \\ \frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}=e^{-\frac{t}{t_{L1}}}-e^{-\frac{t}{t_{L1}}} \\ \frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}=e^{-\frac{t}{t_{L1}}+\frac{t}{t_{L1}}} \\ In\left(\frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}\right)=-\frac{t}{{\mathrm{Ï„}}_{L1}}+\frac{t}{{\mathrm{Ï„}}_{L2}} \\ In\left(\frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}\right)=-\frac{t}{L_1/R_1}+\frac{t}{L_2/R_2} \\ In\left(\frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}\right)=t\left(\frac{R_2}{L_2}-\frac{R_1}{L_1}\right) \\ t=\frac{In\left(\frac{{\mathrm{Ï•}}_{02}}{{\mathrm{Ï•}}_{01}}\right)}{\left(\frac{R_2}{L_2}-\frac{R_1}{L_1}\right)} \\ t=\frac{In\left(1.50\right)}{\left({\displaystyle \frac{30.0\mathrm{Î\copyright }}{3.00×{10}^{-3}H}}-{\displaystyle \frac{25.0\mathrm{Î\copyright }}{5.00×{10}^{-3}H}}\right)} \\ t=81.1×{10}^{-6}s \end{gathered}$$

$t=81.1\mathrm{μ}s$