Question 19.2: Find I1 and I2 in the circuit in Fig. 19.10....

This is not a reciprocal network. We may use the equivalent circuit in Fig. 19.5(b) but we can also use Eq. (19.1) directly. Substituting the given z parameters into Eq. (19.1)

\begin{matrix}V_{1} = z_{11} I_{1}  + z_{12} I_{2}\\ V_{2} = z_{21} I_{1}  + z_{22} I_{2} \end{matrix}\quad \quad     (19.1)

V_{1} = 40I_{1}  +  j20I_{2}                 (19.2.1)

V_{2} = j30I_{1}  +  50I_{2}                 (19.2.2)

Since we are looking for I_{1}   and   I_{2} , we substitute

V_{1} = 100 \underline{/ 0°} ,            V_{2} = -10 I_{2}

into Eqs. (19.2.1) and (19.2.2), which become

100 = 40I_{1}  + j20I_{2}                            (19.2.3)

-10I_{2} = j30I_{1}  +  50 I_{2}          ⇒             I_{1} = j2I_{2}                            (19.2.4)

Substituting Eq. (19.2.4) into Eq. (19.2.3) gives

100 = j80I_{2}  +  j20I_{2}            ⇒              I_{2} = \frac{100}{j100} = -j

From Eq. (19.2.4), I_{1} = j2(-j) = 2  .   Thus , 

I_{1} = 2 \underline{/ 0°} A ,               I_{2} = 1 \underline{/-90°}  A