Respuesta :
Answer:30.08 ms
Explanation:
Given
time Constant [tex]\tau =43.4 ms =\frac{L}{R}[/tex]
Also rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in inductor's magnetic Field
Energy stored in Inductor is [tex]U_L=\frac{1}{2}Li^2[/tex]
rate of Energy storing [tex]\frac{dU_L}{dt}=\frac{1}{2}L\cdot 2i\times \frac{di}{dt}----1[/tex]
Rate of Energy dissipation from resistor i.e. Power is given by
[tex]\frac{dU_R}{dt}=i^2R-----2[/tex]
Equating 1 and 2
[tex]Li\cdot \frac{di}{dt}=i^2R[/tex]
[tex]L(\frac{di}{dt})=R(i)-----3[/tex]
i is given by [tex]i=\frac{V}{R}(1-e^{-\frac{t}{\tau }})[/tex]
[tex]\frac{di}{dt}=\frac{V}{L}e^{-\frac{t}{\tau }}[/tex]
substitute the value of [tex]\frac{di}{dt}[/tex] in 3
[tex]L(\frac{V}{L}e^{-\frac{t}{\tau }})=R\cdot \frac{V}{R}(1-e^{-\frac{t}{\tau }})[/tex]
[tex]e^{-\frac{t}{\tau }}=1-e^{-\frac{t}{\tau }}[/tex]
[tex]2e^{-\frac{t}{\tau }}=1[/tex]
[tex]e^{-\frac{t}{\tau }}=0.5[/tex]
[tex]e^{-\frac{t}{43.4\times 10^{-3}}}=0.5[/tex]
[tex]\frac{t}{43.4\times 10^{-3}}=0.693[/tex]
[tex]t=30.08 ms[/tex]
