Question

In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.

Answer 1

Answer:

The energy is   $U = 18.98 \ J$

Explanation:

From the question we are told that

   The inductor is  $L = 4.74 \ H$

    The resistance of the resistor is $R = 9.33 \ \Omega$

    The voltage of the battery is $V = 26.4 \ V$

Generally the current flowing in the circuit is mathematically represented as

      $I = \frac{V}{R}$

=>   $I = \frac{26.4}{9.33 }$

=>   $I = 2.83 \ A$

Generally the corresponding energy stored in the circuit is  

       $U = \frac{1}{2} * L * I^2$

        $U = \frac{1}{2} * 4.74 * 2.83 ^2$

       $U = 18.98 \ J$