91影视

Faraday鈥檚 law states that a current is induced in a conductor when it is exposed to a time varying magnetic flux. This induced current is driven by a force called electromotive or electromagnetic force. The magnitude of induced emf is given by

$蔚=-L\frac{di}{dt}$

Where L is the inductance of the conductor.

Lenz further explained the direction of this induced current. According to lenz, the direction of induced current will be such that the magnetic field created by the induced current opposes the changing magnetic field which caused its induction.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. An inductor stores the energy in form of magnetic field.

A capacitor is a device consisting of two conductors in close proximity that are used to store electrical energy. A capacitor stores energy in form of electric field.

An LC-circuit can store energy while oscillating at its natural resonant frequency. For such circuits, the energy in the circuit shifts between electric and magnetic fields. The angular frequency of this oscillation is given by

$蝇=2\mathrm{蟺}f=\frac{1}{\sqrt{LC}}$

Apply kirchhoff鈥檚 rule on the bottom loop, we get

$\begin{array}{r}蔚鈭�v_x鈭�v_{cb}=0\\ 蔚鈭�i{R}_0鈭�iR鈭�L\frac{di}{dt}=0\\ L\frac{di}{dt}=蔚鈭�i\left(R_0+R\right)\\ \frac{di}{鈭�i+蔚/\left(R_0+R\right)}=\left(\frac{R_0+R}{L}\right)dt\end{array}$

Integrating both sides from t_i = 0 to t_f = t sec

$\begin{array}{r}\ln 鈦�\left(\frac{鈭�i_0+蔚/\left(R_0+R\right)}{蔚/\left(R_0+R\right)}\right)=鈭�\left(\frac{R_0+R}{L}\right)t\\ \frac{鈭�i_0+蔚/\left(R_0+R\right)}{蔚/\left(R_0+R\right)}=e^{鈭�\left(\frac{R_0+R}{L}\right)t}\\ i_0=\frac{蔚}{\left(R_0+R\right)}\left(1鈭�e^{鈭�\left(\frac{R_0+R}{L}\right)t}\right)\end{array}$

Using this current variation, voltage drop across inductor is given by

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

Therefore, the variation of voltage across solenoid is represented using above equation.

As the voltage drop across inductor right after closing the switch is 50 volts and there鈥檚 no drop of voltage anywhere else in the circuit.

Therefore, the EMF of the battery is 50 volts.

As we know the voltage drop across inductor reaches to its minimum when a long time has passed, which is given by, V_L(min) = 20 V, voltage drop across resistor is emf - V_L(min) = 50 - 20 V = 30 V.

At this point, using ohm鈥檚 law current through the resistor is given by, i = V/R = 3A

Therefore, the voltage drop across resistor is 30 volts and the current through resistor is 3 amperes.

Using ohm鈥檚 law to get the resistance of coil is

$R_L=\frac{V_L}{l}=\frac{20V}{3A}=6.7惟$

Therefore, the resistance of the solenoid is 6.7 ohms.

By step 2, voltage drop across coil is given by

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

Put all the value calculated in the equation, V_L = 31 volts.

From the graph, at V_L = 31 volts, time constant is 2.3 ms. Thus, self-inductance of the coil is given by

$\begin{array}{r}L=蟿\left(R+R_L\right)\\ =\left(2.3\mathrm{ms}\right)(10+6.7)\mathrm{惟}\\ =38.41\mathrm{mH}\end{array}$

Therefore, the self-inductance of the coil is 38.41 milli-henry.