Consider the circuit in Exercise 30.23. When the current has reached its final steady-state value, how much energy is stored in the inductor? What is the rate at which electrical energy is being dissipated in the resistance of the inductor? What is the rate at which the battery is supplying electrical energy to the circuit?

The current in RL circuit at any time is given by

$i=\frac{E}{R}\left(1-e^{\left(R/L\right)t}\right)$

The current in RL circuit at steady state $(t\to \infty )$

In steady state the current in RL circuit becomes constant and does not change with time.

So, the current in RL circuit at steady state is given by

$i=\frac{E}{R}$

Where, is the E emf or voltage supply and R is the resistance.

The energy stored in inductor at steady state

The energy stored in inductor at steady state is given by

$\mathrm{U}=\frac{1}{2}L{\mathrm{i}}^2$

Where L is the inductance , i is current in the circuit and U is the energy stored in inductor

The rate at which electrical energy is being dissipated in the resistance

The rate at which electrical energy is being dissipated in the resistance of the inductor in steady state is given by

$$\begin{gathered} P_R=i^{2}R \\ {P}_R={\left(\frac{E}{R}\right)}^{2}R \end{gathered}$$

Where,${\mathrm{P}}_R$ is the power dissipated in the resistance .

The rate at which battery is supplying electrical energy to the circuit

The rate at which the battery is supplying electrical energy to the circuit in steady state is given by

${\mathrm{P}}_E=\mathrm{iE}$

Where${\mathrm{P}}_E$ is the power supplied by battery.

Given : Inductance of inductor (L) = 2.50

Resistance $R=8.00\Omega$

EMF E = 6.00V

Using

$$\begin{gathered} U=\frac{1}{2}L{i}^2 \\ U=\frac{1}{2}L{\left(\frac{E}{R}\right)}^2 \end{gathered}$$

Put the value of constants in above equation

role="math" localid="1668239684420" $$\begin{gathered} U=\frac{1}{2}(2.50H){\left(\frac{6.00}{8.00\Omega }\right)}^2 \\ U=0.703J \end{gathered}$$

Thus, the energy stored in the inductor in steady state is .

Using

$$\begin{gathered} P_R=i^{2}R \\ {P}_R={\left(\frac{E}{R}\right)}^{2}R \end{gathered}$$

Put the values of constants in above equation

$$\begin{gathered} P_R={\left(\frac{6.00V}{8.00\Omega }\right)}^2(8.00\Omega ) \\ {P}_R=4.50W \end{gathered}$$

Thus, the rate at which electrical energy is being dissipated in the resistance of the inductor is 4.50 W.

Using

$$\begin{gathered} P_E=iE \\ {P}_E=\frac{E}{R}\left(E\right) \end{gathered}$$

Now put the values of constants in above equation

$$\begin{gathered} P_E=\frac{6V\times 6V}{8.00\Omega } \\ {P}_E=4.50W \end{gathered}$$

Thus, the rate at which the battery is supplying electrical energy to the circuit is 4.50W.