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.

Respuesta :

Answer:

The energy is   [tex]U =  18.98 \  J [/tex]

Explanation:

From the question we are told that

   The inductor is  [tex]L  =  4.74 \ H[/tex]

    The resistance of the resistor is [tex]R =  9.33 \  \Omega[/tex]

    The voltage of the battery is [tex]V =  26.4 \  V[/tex]

Generally the current flowing in the circuit is mathematically represented as

      [tex]I =  \frac{V}{R}[/tex]

=>   [tex]I =  \frac{26.4}{9.33 }[/tex]

=>   [tex]I =  2.83 \ A[/tex]

Generally the corresponding energy stored in the circuit is  

       [tex]U =  \frac{1}{2} * L  *  I^2[/tex]

        [tex]U =  \frac{1}{2} *  4.74  *  2.83 ^2[/tex]

       [tex]U =  18.98 \  J [/tex]