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 = lR

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

${\underset{}{\mathbf{鈭�}}}^{\mathbf{V}}\mathbf{=}\mathbf{0}$

Use Kirchhoff鈥檚 laws in the bottom loop to get the equation

$\begin{array}{r}蔚鈭�v_{ax}鈭�v_{cb}=0\\ 蔚鈭�i{R}_0鈭�iR鈭�L\frac{di}{dt}=0\end{array}$

Rearrange to get the equation

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

Integrate both sides

$\ln 鈦�\left(\frac{鈭�i_0+\frac{蔚}{R+R_0}}{\frac{蔚}{R+R_0}}\right)=鈭�\left(\frac{R+R_0}{L}\right)t$

Take exponents on both sides

$\frac{鈭�i_0+\frac{蔚}{R+R_0}}{\frac{蔚}{R+R_0}}=e^{鈭�\left(\frac{R+R_0}{L}\right)t}$

Rearrange and we get

$i_0=\frac{蔚}{R+R_0}\left(1鈭�e^{鈭�\left(\frac{R+R_0}{L}t\right)}\right)$

Use Ohm鈥檚 law to get the voltage drop across the Inductor

$\begin{array}{c}{V}_L=i_0{R}_L\\ V_L=\frac{蔚R_L}{R+R_0}\left(1鈭�e^{鈭�\left(\frac{R+R_0}{L}t\right)}\right.\end{array}$

We are given that the voltage across the inductor is 50V just after the switch is closed. In this case, no voltage drop occurs in the circuit which means the voltage of the inductor equals to the emf of the battery

emf = 50V

At time$t=鈭�$ the voltage across the inductor reaches its minimum value which is given by$V_{L\left(min\right)}=20V$. So,Using the conservation law, we get the voltage of the resistance by

$\begin{array}{r}{V}_R=emf鈭�V_{L(min)}\\ V_R=50\mathrm{V}鈭�20\mathrm{V}\end{array}$

Use Ohm鈥檚 law to get current through$R=10惟$.

$\begin{array}{c}{i}_R=\frac{V_R}{R}=\frac{30V}{10\mathrm{惟}}\\ i_R=3A\end{array}$

The voltage across Resistance is$V_R=30V$ and current is$i_R=3A$

We get the voltage of the inductor to be 20V, So we use the Ohm鈥檚 Law to get the resistance of the coil

$\begin{array}{c}{R}_L=\frac{V_L}{i}=\frac{20\mathrm{V}}{3\mathrm{A}}\\ R_L=6.7\mathrm{惟}\end{array}$

The resistance of inductor is$6.7惟$

Use

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

Substitute the value of the variables and$-\left(\frac{R+R_0}{L}\right)t$ to be 1 to get the value of$V_L$

$V_L=31V$

Which corresponds the time constant to berole="math" localid="1668250939109" $蟿=2.3ms$

Hence

$L=蟿\left(R+R_0\right)$

Plug the values of the variables and we get

L = 38.41mH

The inductance of the inductor is L = 38.41mH