Q86P - Fundamentals Of Physics

Question

In Fig. 30-75a, switch S has been closed on A long enough to establish a steady current in the inductor of inductance ${\mathit{L}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{mH}$and the resistor of resistance ${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}$. Similarly, in Fig. 30-75b, switch S has been closed on A long enough to establish a steady current in the inductor of inductance ${\mathit{L}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{mH}$and the resistor of resistance ${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{\Omega }$. The ratio $\frac{{\mathbf{\varphi }}_{\mathbf{02}}}{{\mathbf{\varphi }}_{\mathbf{01}}}$of the magnetic flux through a turn in inductor 2 to that in inductor 1 is  . At time t=0 , the two switches are closed on B. At what time t is the flux through a turn in the two inductors equal?

Step-by-Step Solution

Answer

The time at which the flux through a turn in two inductors equals is $t=81.1 \mathrm{\mu s}$

Step 1: Given

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

ii) The resistor of resistance is $R_1=25.0\Omega =20\times {10}^{-3}\Omega$ 

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

iv) The resistor of resistance is $R_2=30.0\Omega$ 

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

Step 2: Understanding the concept

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

 

Formulae:

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

 

Step 3: Calculate the time at which the flux through a turn in two inductors equals

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,

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

 

The expression of the inductive time constant is

$t=\frac{L}{R}$ 

According to the condition,

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

$t=81.1\mu s$