Respuesta :
Answer:
t = 186.2 μs
Explanation:
Current in LR series circuit
[tex]I(t) = I_{s}( 1 - e^{-Rt/L)}[/tex]----(1)
steady current = I_{s} = V/R
time constant = τ =[tex]L/R =4.3 * 10^{-3} / 16\\[/tex]
= 0.268 ms
magnetic energy stored in coil = [tex]U_{L} = \frac{1}{2}LI^{2}[/tex]
rate at which magnetic energy stored in coil= [tex]\frac{d}{dt}U_{L} =\frac{d}{dt} \frac{1}{2}LI^{2} \\ = LI\frac{dI}{dt}\\[/tex]----(2)
rate at which power is dissipated in R:
[tex]P = I^{2}R[/tex]---(3)
To find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil equate (2) and (3)
[tex]I^{2}R=LI \frac{dI}{dt}[/tex]
[/tex]I=\frac{L}{R}\frac{dI}{dt}[/tex]----(4)
differentiating (1) w.r.to t
[tex]I(t)=I_{f} (1-e^{\frac{Rt}{L} })[/tex]
[tex]\frac{dI}{dt} = I_{f}\frac{d}{dt}(1-e^{\frac{-Rt}{L} } )[/tex]
[tex]\frac{dI}{dt}= I_{f}(-\frac{R}{L} e^{\frac{-Rt}{L} } )\\[/tex]---(5)
substituting (5) in (4)
[tex]I=I_{f}e^{-\frac{Rt}{L} }[/tex]----(6)
equating (1) and (6)
[tex]I_{f}( 1- e^{-\frac{Rt}{L} } ) = I_{f}e^{-\frac{Rt}{L} }[/tex]
[tex]1 - e^{-\frac{Rt}{L} } = e^{-\frac{Rt}{L} }[/tex]
[tex]\frac{1}{2}= e^{-\frac{Rt}{L} }[/tex]
[tex]t= -\frac{L}{R}ln\frac{1}{2}[/tex]
L= 4.3 mH
R= 16 Ω
t = 186.2 μs
