# No initial current in the inductor, switch in open state find: (a) Immediately after Close, I_1, I_2, I_3, & V_L? (b) Close long I_1, I_2, I_3, & V_L? (c) Immediately after Open, I_1, I_2, I_3, & V_L? (d) Open Long, I_1, I_2, I_3, & V_L?

Nov 29, 2016

Considering two independent currents ${I}_{1}$ and ${I}_{2}$ with two independents loops we have

loop 1) $E = {R}_{1} {I}_{1} + {R}_{1} \left({I}_{1} - {I}_{2}\right)$
loop 2) ${R}_{2} {I}_{2} + L {\dot{I}}_{2} + {R}_{1} \left({I}_{2} - {I}_{1}\right) = 0$ or

$\left\{\begin{matrix}2 {R}_{1} {I}_{1} - {R}_{1} {I}_{2} = E \\ - {R}_{1} {I}_{1} + \left({R}_{1} + {R}_{2}\right) {I}_{2} + L {\dot{I}}_{2} = 0\end{matrix}\right.$

Substituting ${I}_{1} = \frac{E - {R}_{1} {I}_{2}}{2 {R}_{1}}$ into the second equation we have

$E + \left({R}_{1} + 2 {R}_{2}\right) {I}_{2} + 2 L {\dot{I}}_{2} = 0$ Solving this linear differential equation we have

${I}_{2} = {C}_{0} {e}^{- \frac{t}{\tau}} + \frac{E}{{R}_{1} + 2 {R}_{2}}$ with $\tau = \frac{2 L}{{R}_{1} + 2 {R}_{2}}$

The constant ${C}_{0}$ is determined according to the initial conditions.

${I}_{2} \left(0\right) = 0$ so

$0 = {C}_{0} + \frac{E}{{R}_{1} + 2 {R}_{2}}$

Substituting ${C}_{0}$ we have

${I}_{2} = \frac{E}{{R}_{1} + 2 {R}_{2}} \left(1 - {e}^{- \frac{t}{\tau}}\right)$

Now we can answer the items.

a) ${I}_{2} = 0 , {I}_{1} = \frac{10}{8} , {V}_{L} = \frac{10}{8} 4$
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) ${I}_{1} = {I}_{2} = {V}_{L} = 0$