Answer:
a) i=-16A
b) E=1.28 J
c) T=2.5 m
d) i(t)=-16e^(-400*t)
e) E=1.26 J
Explanation:
The circuit is shown in the attached diagram:
a) at t=0, the circuit is in steady state and we have the following:
[tex](\frac{V-100}{1})+\frac{V}{20}+\frac{V}{5}=0\\ V(1+\frac{1}{20}+\frac{1}{5})=100[/tex]
Clearing V:
V=80 V
[tex]i=\frac{V}{S}=16A\\ i=-16A[/tex]
b) the initial energy stored is equal to:
[tex]E=\frac{1}{2}Li^{2}=\frac{1}{2}*10*16^{2}=1.28 J[/tex]
c) at t>0
[tex]T=\frac{L}{R}=\frac{10}{4}=2.5 m[/tex]
d) if t=∝, i=0
[tex]i(t)=i(\alpha )+(i(o)-i(\alpha ))e^\frac{-t}{T}=0+(16-0)e^{\frac{-t}{2.5x10^{-5} } }[/tex]
for t≥0
[tex]i(t)=-16e^{-400t}[/tex]
e) the energy disipation in 4Ω is equal to:
[tex]E=\int\limits^5_0 {i^{2}*4dt } \,\\ E=\int\limits^5_0 {(-16e^{-400t})^{2} *4dt } \,\\ E=1024\int\limits^5_0 {e^{-800t} } \, dt \\E=\frac{1024}{-800}e^{-800t}|(5-0)=1.26 J[/tex]
