Find the voltage:

Solution:

$ \begin{aligned} \\ &\text { Apply KVL to loop (1) }\\\\ &120 \lfloor a-20(a+b)-j 20(b)-j 15(a)=0\\\\ &-(20+j 15) a-(20+j 20) b=-120 \lfloor 0......(1)\\\\ &\text { Apply KVL to second loop }\\\\ &\begin{aligned} \\ &-j 10(a)-j 15(b)-15(a)+j 20(b)+j 15(a)=0 \\\\ &-(15-j 5) a+(j 5) b=0 \\ \end{aligned}\\\\ &\Delta=\left|\begin{array}{cc} \\ -(20+j 15) & -(20+j 20) \\ -(15-j 5) & (j 5) \end{array}\right|\\\\ &=-(20+j 15)(j 5)-(20+j 20)(15-j 5)\\\\ &=-325-300 j=442.29-137.29 \end{aligned}\\ $

$ \begin{aligned} \\ &\Delta_a=\left|\begin{array}{cc} \\ -120 \lfloor 0 & -(20+j 20) \\\\ 0 & (j 5) \end{array}\right|\\\\ &=-(120)(j 5)=-600 j=600 \text { $\lfloor$ - 90 }\\\\ &\therefore a=\frac{\Delta a}{\Delta}=\frac{600 L-90}{442.29 \lfloor-137.29}=1.356447 .29 \text { Amp }\\\\ &\text { Hence voltage across } 15 \Omega \text { resistor is }\\\\ &\begin{gathered}\\ V_{15}=15(a)=15(1.356 \lfloor 47.29) \\\\ V_{15}=20.3447 .29 \text { Volts }\\ \end{gathered}\\ \end{aligned}\\ $