Question 7.3: Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the c......

There are two ways we can solve this problem. One way is to obtain the equivalent resistance at the inductor terminals and then use Eq. (7.20). The other way is to start from scratch by using Kirchhoff’s voltage law. Whichever approach is taken, it is always better to first obtain the inductor current.

i(t) = I_{0} e^{-t/\tau}                (7.20)

■ METHOD 1 The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with v_{o} = 1 V  at the inductor terminals a–b, as in Fig. 7.14(a). (We could also insert a 1-A current source at the terminals.) Applying KVL to the two loops results in

2(i_{1} – i_{2} ) + 1 = 0    ⇒   i_{1} – i_{2} = – \frac{1}{2}          (7.3.1)

6i_{2} – 2i_{1} – 3i_{1} = 0      ⇒     i_{2} = \frac{5}{6} i_{1}           (7.3.2)

Substituting Eq. (7.3.2) into Eq. (7.3.1) gives

i_{1} = -3 A ,     i_{o} = – i_{1} = 3 A

Hence ,

R_{eq} = R_{Th} = \frac{v_{o}}{i_{o}} = \frac{1}{3} Ω

The time constant is

\tau = \frac{L}{R_{eq}} = \frac{\frac{1}{2}}{\frac{1}{3}} = \frac{3}{2} s

Thus, the current through the inductor is

i(t) = i(0) e^{-t/\tau} = 10 e^{-(2/3)t} A ,            t > 0

■ METHOD 2 We may directly apply KVL to the circuit as in Fig. 7.14(b). For loop 1,

\frac{1}{2} \frac{di_{1}}{dt} + 2(i_{1} – i_{2} ) = 0

or

\frac{di_{1}}{dt} + 4i_{1} – 4i_{2} = 0         (7.3.3)

For loop 2 ,

6i_{2} – 2i_{1} – 3i_{1} = 0    ⇒   i_{2} = \frac{5}{6} i_{1}        (7.3.4)

Substituting Eq. (7.3.4) into Eq. (7.3.3) gives

\frac{di_{1}}{dt} + \frac{2}{3} i_{1} = 0

Rearranging terms,

\frac{di_{1}}{i_{1}} = – \frac{2}{3} dt

Since i_{1} = i   we may replace i_{1}   with i and integrate: 

ln i \mid^{i(t)}_{i(0)} = – \frac{2}{3}t\mid^{t}_{0}

or

ln \frac{i(t)}{i(0)} = – \frac{2}{3} t

Taking the powers of e, we finally obtain

i(t) = i(0)e^{-(2/3)t} = 10e^{-(2/3)t} A ,       t > 0 

which is the same as by Method 1.
The voltage across the inductor is

v = L \frac{di}{dt} = 0.5 (10)(-\frac{2}{3}) e^{-(2/3)t} = – \frac{10}{3} e^{-(2/3)t} V

Since the inductor and the  2-Ω resistor are in parallel,

i_{x}(t) = \frac{v}{2} = -1.6667e^{-(2/3)t} A ,        t> 0