Solving bu engineering analysis ?

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Answer 1 · 2018-01-07T21:06:06

See below.

With the switch open we have

$i_{10} = i_{20} = \frac{E}{R_{1} + R_{2}}$ $i_(10) = i_(20) = E/(R_1+R_2)$

After closing the switch

${(E = i_1 R_1+(i_1-i_2)R_3),(i_2 R_2+d/(dt)i_2 L+(i_2-i_1)R_3=0):}$ or

${((R_1+R_3)i_1-R_3i_2 = E),(-R_3 i_1+(R_2+R_3+d/(dt)L)i_2 = 0):}$

substituting for $i_2$

$[((R_1+R_3)/R_3)(R_2+R_3+d/(dt)L)-R_3]i_1=((R_2+R_3)/R_3) E$

now calling

$eta = ((R_1+R_3)/R_3)$
$lambda = eta(R_2+R_3)-R_3$ and
$mu = etaL$

we have

$lambda i_1 + mu d/(dt) i_1 = eta E$ with solution

$i_1 = eta/lambda E + C_0 e^(-lambda/mu t)$ but at $t=0$ we have $i_1 = i_(10)$ then

$i_(10) = eta/lambda E +C_0 rArr C_0 = i_(10)-eta/lambda E$ and then

$i_1 = eta/lambda E+(i_(10)-eta/lambda E)e^(-lambda/mu t)$ or

$i_1 = i_(10)e^(-lambda/mu t)+eta/lambda E(1-e^(-lambda/mu t))$

With the same procedure we obtain $i_2$ and is left as an exercise.