Question 7.12: Find i(t) in the circuit of Fig. 7.51 for t > 0 . Assume......

When t < 0 ,  the 3-Ω  resistor is short-circuited, and the inductor acts like a short circuit. The current through the inductor at t = 0^{-}   (i.e., just before t = 0  ) is

i(o^{-} ) = \frac{10}{2} = 5 A

Since the inductor current cannot change instantaneously,

i(0) = i(0^{+}) = i(0^{-}) = 5 A

When t > 0 ,  the switch is open. The  2-Ω and 3-Ω  resistors are in series, so that

i(∞) = \frac{10}{2 + 3} = 2 A

The Thevenin resistance across the inductor terminals is

R_{Th} = 2 + 3 = 5 Ω

For the time constant,

\tau = \frac{L}{R_{Th}} = \frac{\frac{1}{3}}{5} = \frac{1}{15} s

Thus,

i(t0 = i ( ∞) + [i(0) – i(∞) ] e^{-t/\tau}

= 2 + (5 – 2) e^{-15t} = 2 + 3 e^{-15t} A ,          t > o

Check: In Fig. 7.51, for  t > 0 , KVL must be satisfied; that is,

10 = 5i + L \frac{di}{dt}

5i + L \frac{di}{dt} = [ 10 + 15 e^{-15t} ] + [ \frac{1}{3} (3) (-15) e^{-15t} ] = 10

This confirms the result.