91影视

Ohm鈥檚 law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

V = IR

According to Kirchhoff鈥檚 Lawthe sum of the voltages around the closed loop is equal to null.

$\underset{}{鈭�}V=0$

When the switch is just closed the initial current is zero i.e$i_0=0A$ and in this case the voltage across ab will be zero $V_{ab}=0$.

And as no voltage drop occurs therefore the voltage across bc is the same of the emf of the battery$V_{bc}=36V$

Hence the voltmeters read$V_{ab}=0,V_{bc}=36V$ and the current is $i_0=0$.

After a long time of closing of the switch, the voltage across the inductor is zero, so use the ohm鈥檚 law to get$i_0$.

$\begin{array}{r}{i}_0=\frac{蔚}{R_{eq}}\\ i_0=\frac{36\mathrm{V}}{(150\mathrm{惟}+30\mathrm{惟})}\\ i_0=0.18\mathrm{A}\end{array}$

Now use this value to get the voltage across$V_{ab}$ and$V_{bc}$ by

$V_{ab}=i_0{R}_0=\left(0.18\mathrm{A}\right)\left(50\mathrm{惟}\right)=9\mathrm{V}$

And

$V_{bc}=i_{0}R=(0.18A)\left(150惟\right)=27V$

Hence The voltmeters read$V_{ab}=9,V_{bc}=27V$ and the current is $i_0=0.18A$.

Use Kirchhoff鈥檚 rule in the bottom loop and get the equation

$\begin{array}{r}蔚鈭�V_{ac}鈭�V_{cb}=0\\ 蔚鈭�i_0{R}_0鈭�i_{0}R鈭�L\frac{di}{dt}=0\end{array}$

Rearranging

$\begin{array}{r}L\frac{di}{dt}=蔚鈭�i_0\left(R_0+R\right)\\ \frac{di}{鈭�i+\frac{蔚}{\left(R_0+R\right)}}=\frac{\left(R_0+R\right)}{L}dt\end{array}$

Integrate both sides

$\ln 鈦�\left(\frac{鈭�i+\frac{蔚}{\left(R_0+R\right)}}{\frac{蔚}{\left(R_0+R\right)}}\right)=鈭�\frac{\left(R_0+R\right)}{L}dt$

Take exponent on both sides

$\frac{鈭�i+\frac{蔚}{\left(R_0+R\right)}}{\frac{蔚}{\left(R_0+R\right)}}=e^{鈭�\frac{\left(R_0+R\right)}{L}dt}$

Rearrange to get

$i_{0=}\frac{蔚}{\left(R_0+R\right)}\left(1鈭�e^{鈭�\frac{\left(R_0+R\right)}{L}t}\right)$

Plug in the values$R,R_0,Land蔚$ of to get

$i_0=0.180A\left(1鈭�e^{鈭�\frac{t}{0.02s}}\right)$

Hence the current is$i_0=0.180A\left(1鈭�e^{鈭�\frac{t}{0.02s}}\right)$