Question

The switch in the circuit has been closed for a long time and is opened at t = 0.

Answer 1

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:

$(\frac{V-100}{1})+\frac{V}{20}+\frac{V}{5}=0\\ V(1+\frac{1}{20}+\frac{1}{5})=100$

Clearing V:

V=80 V

$i=\frac{V}{S}=16A\\ i=-16A$

b) the initial energy stored is equal to:

$E=\frac{1}{2}Li^{2}=\frac{1}{2}*10*16^{2}=1.28 J$

c) at t>0

$T=\frac{L}{R}=\frac{10}{4}=2.5 m$

d) if t=∝, i=0

$i(t)=i(\alpha )+(i(o)-i(\alpha ))e^\frac{-t}{T}=0+(16-0)e^{\frac{-t}{2.5x10^{-5} } }$

for t≥0

$i(t)=-16e^{-400t}$

e) the energy disipation in 4Ω is equal to:

$E=\int\limits^5_0 {i^{2}*4dt } \,\\ E=\int\limits^5_0 {(-16e^{-400t})^{2} *4dt } \,\\ E=1024\int\limits^5_0 {e^{-800t} } \, dt \\E=\frac{1024}{-800}e^{-800t}|(5-0)=1.26 J$