Case A

If there is a voltage source Vsk in the branch K having impedance Zk & carrying Ik then

$V_k=Z_k\ ik-Vsk\ \left(k=1,2,…..b\right) $

In matrix form  $V_b=Z_bI_b-V_s$

$Z_b=branch\ impedance\ matrix,\ I_b=column\ vector\ of\ branch\ currents,\ $

$V_s=column\ vector\ of\ source\ voltages\ $

∴ KVL equation can be written as

$BV_b=0 $

$B\left(Z_bI_b-Vs\right)=0 $

$BZ_bI_b=BVs $

$Also\ I_b=B^TI_L $

$BZ_bB^TI_L=BVs $

$ZI_L=E\ \ where\ E=BVs,\ Z=BZ_BB^T $

Z= loop impedance matrix

Case B

If there is voltage source in series with an impedance & current source in parallel

$I_k=\dfrac{V_k+V_{sk}}{z_k}-isk\ $

$V_k=Z_ki_k+Z_ki_{sk}-Vsk\ $

In matrix form, $V_b=Z_bI_b+Z_bI_s-V_s$

$KVL\ equation\ is\ BV_b=0 $

$BV_b=B\left(Z_bI_b+Z_bI_s-V_s\right)=0 $

$BZ_bI_b=BVs-BZ_bI_s $

$but\ I_b=B^TI_L $

$\therefore{}BZ_bB^TI_1=BVs-BZ_bI_s $

$\therefore{}ZI_L=BVs-BZ_bI_s $

$where\ z=BZ_bB^T=loop\ impedance\ matrix\ $

$ This\ is\ generalised\ KVL\ equation\ $