91影视

When the current is supplied to a circuit in a magnetic field, then it moves against the direction of the back emf.

The work done by a charge in moving against the back emf of the circuit is described as the 鈥榚nergy in the magnetic field鈥�.

Using Ohm鈥檚 law, the formula for the initial current of the circuit is given by,

$I_0=\frac{{蔚}_0}{R}$

The formula for the Induced emf in the circuit is given by,

$$\begin{gathered} -L\frac{dI}{dt}=IR \\ \frac{dI}{dt}=-\frac{R}{L}I \end{gathered}$$

Then the expression for the current in the circuit as a function of the subsequent time is given by,

$$\begin{gathered} I=I_0{e}^{\frac{-Rt}{L}} \\ I\left(t\right)=\frac{{蔚}_0}{R}{e}^{\frac{-Rt}{L}} \end{gathered}$$

Hence, the current at any subsequent time is$I\left(t\right)=\frac{{蔚}_0}{R}{e}^{\frac{-Rt}{L}}$.

The formula for the power due to the resistance of the circuit is given by,

$$\begin{gathered} P=I^{2}R \\ P={\left(\frac{{蔚}_0}{R}e\frac{-Rt}{L}\right)}^{2}R \\ P={\left(\frac{{蔚}_0}{R}\right)}^2{e}^{\frac{-2Rt}{L}}R \\ P=\left(\frac{{{蔚}_0}^2}{R^2}\right)e^{\frac{-2Rt}{L}}R \end{gathered}$$

Rewrite the equation as:

$P=\left(\frac{{{蔚}_0}^2}{R}\right)e^{\frac{-2Rt}{L}}$

Similarly, the formula for the power required by the charge to move against the emf is given by,

$$\begin{gathered} P=\frac{dW}{dt} \\ dW=Pdt \\ dW=\left(\frac{{{蔚}_0}^2}{R}\right)e^{\frac{-2Rt}{L}}dt \end{gathered}$$

Here, W is the work done or the energy delivered to the resistor.

Integrating both sides,

$$\begin{gathered} W=\left(\frac{{{蔚}_0}^2}{R}\right){鈭�}_0^{鈭�}{e}^{\frac{-2Rt}{L}}dt \\ W=\left(\frac{{{蔚}_0}^2}{R}\right){\left(-\frac{L}{2R}{e}^{\frac{-2Rt}{L}}\right)}_0^{鈭�} \\ W=\left(\frac{{{蔚}_0}^2}{R}\right)\left(0+\frac{L}{2R}\right) \\ W=\frac{1}{2}L{\left(\frac{{蔚}_0}{R}\right)}^2 \end{gathered}$$

Hence, the energy delivered to the resistor is$\frac{1}{2}L{\left(\frac{{蔚}_0}{R}\right)}^2$.

Using the energy formula, the expression for theenergy originally stored in the inductor is given by,

$$\begin{gathered} W_0=\frac{1}{2}L{I}_0^2 \\ {W}_0=\frac{1}{2}L{\left(\frac{{蔚}_0}{R}\right)}^2 \end{gathered}$$

Comparing the energydelivered to the resistor with the energy originally stored in the inductor,

$W=W_0$

Hence, the energy delivered to the resistor is equal to the energy originally stored in the inductor.