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

The energy is
`U =  18.98 \  J`

Step-by-step 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 =  (V)/(R)`

=>
`I =  (26.4)/(9.33 )`

=>
`I =  2.83 \ A`

Generally the corresponding energy stored in the circuit is

`U =  (1)/(2) * L  *  I^2`

`U =  (1)/(2) *  4.74  *  2.83 ^2`

`U =  18.98 \  J`