The switch in Fig. 6.86 has been in position A for a long time. At t = 0, the switch moves from position A to B. The switch is a make-before-break type so that there is no interruption in the inductor current. Find:

(a) i(t) for t > 0,

(b) v just after the switch has been moved to position B,

(c) v(t) long after the switch is in position B.

Question

The switch in Fig. 6.86 has been in position A for a long time. At t = 0, the switch moves from position A to B. The switch is a make-before-break type so that there is no interruption in the inductor current. Find:

(a) i(t) for t > 0,

(b) v just after the switch has been moved to position B,

(c) v(t) long after the switch is in position B.

(a) i(t) for t > 0,

(b) v just after the switch has been moved to position B,

(c) v(t) long after the switch is in position B.

Accepted Answer

(a) When the switch is in position A,

i= –6 = i(0)

When the switch is in position B,

i(∞)=12/4=3, τ=L/R=1/8

i(t)=i(∞)+[i(0)-i(∞)][latex]e^{-t/τ}[/latex]=3-9[latex]e^{-8t}[/latex] A

(b) -12 + 4i(0) + v=0, i.e. v=12 – 4i(0) = 36 V

(c) At steady state, the inductor becomes a short circuit so that v= 0 V

Question 6.64: The switch in Fig. 6.86 has been in position A for a long ti...

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