Answer to Question #120343 in Electric Circuits for Dave Matthew

Question #120343
https://prnt.sc/su2cfr
1
Expert's answer
2020-06-08T10:29:53-0400

With reference to the image :

For the given circuit,τ=LR1=L15=12    L=7.5\tau=\frac{L}{R_1}=\frac{L}{15}=\frac{1}{2}\implies L=7.5

And at t=0,W=12Lio2=2.22 J    io=4.44L=t=0,W=\frac{1}{2}Li_o^2=2.22\ J\implies i_o=\sqrt{\frac{4.44}{L}}= 4.447.5=0.77A\sqrt{\frac{4.44}{7.5}}=0.77A

At t=0,t=0, inductance offer infinite resistance,So,no current pass through it.

io=ER2    R2=Eio=100.77=13Ωi_o=\frac{E}{R_2}\implies R_2=\frac{E}{i_o}=\frac{10}{0.77}=13 \Omega

And i=io(1et/τ)=0.77(1e2t)i=i_o(1-e^{t/\tau})=0.77(1-e^{2t})


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