Respuesta :
Given Information:
Resistance = R = 14 Ω
Inductance = L = 2.3 H
voltage = V = 100 V
time = t = 0.13 s
Required Information:
(a) energy is being stored in the magnetic field
(b) thermal energy is appearing in the resistance
(c) energy is being delivered by the battery?
Answer:
(a) energy is being stored in the magnetic field ≈ 219 watts
(b) thermal energy is appearing in the resistance ≈ 267 watts
(c) energy is being delivered by the battery ≈ 481 watts
Explanation:
The energy stored in the inductor is given by
[tex]U = \frac{1}{2} Li^{2}[/tex]
The rate at which the energy is being stored in the inductor is given by
[tex]\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1[/tex]
The current through the RL circuit is given by
[tex]i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })[/tex]
Where τ is the the time constant and is given by
[tex]\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16[/tex]
[tex]i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}[/tex]
Therefore, eq. 1 becomes
[tex]\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})[/tex]
At t = 0.13 seconds
[tex]\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts[/tex]
(b) thermal energy is appearing in the resistance
The thermal energy is given by
[tex]P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts[/tex]
(c) energy is being delivered by the battery?
The energy delivered by battery is
[tex]P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts[/tex]
