No initial current in the inductor, switch in open state find: (a) Immediately after Close, $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ? (b) Close long $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ? (c) Immediately after Open, $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ? (d) Open Long, $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ?

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No initial current in the inductor, switch in open state find: (a) Immediately after Close, $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ? (b) Close long $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ? (c) Immediately after Open, $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ? (d) Open Long, $I_{1}, I_{2}, I_{3}, & V_{L}$ $I_1, I_2, I_3, & V_L$ ?

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Answer 1 · 2016-11-29T21:15:18

Considering two independent currents $I_{1}$ $I_1$ and $I_{2}$ $I_2$ with two independents loops we have

loop 1) $E = R_{1} I_{1} + R_{1} (I_{1} - I_{2})$ $E=R_1I_1+R_1(I_1-I_2)$
loop 2) $R_{2} I_{2} + L {. I}_{2} + R_{1} (I_{2} - I_{1}) = 0$ $R_2I_2+L dot I_2+R_1(I_2-I_1)=0$ or

${(2R_1 I_1-R_1I_2=E),(-R_1I_1+(R_1+R_2)I_2+L dot I_2=0):}$

Substituting $I_1=(E-R_1I_2)/(2R_1)$ into the second equation we have

$E+(R_1+2R_2)I_2+2L dot I_2=0$ Solving this linear differential equation we have

$I_2=C_0e^(-t/tau)+E/(R_1+2R_2)$ with $tau=(2L)/(R_1+2R_2)$

The constant $C_0$ is determined according to the initial conditions.

$I_2(0)=0$ so

$0=C_0+E/(R_1+2R_2)$

Substituting $C_0$ we have

$I_2=E/(R_1+2R_2)(1-e^(-t/tau))$

Now we can answer the items.

a) $I_2=0,I_1=10/8,V_L=10/8 4$
b) $I_2=10/(4+2 cdot8),I_1=?, V_L=0$
c) $I_2=?,I_1=0,V_L=?$ we let those answers to the reader
d) $I_1=I_2=V_L=0$

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