Question 3.26: Two single-phase furnaces A and B are supplied at 100 V by m...

With reference to Fig.3.64(a)
\frac{N_{1} }{N_{2} } =\frac{6600}{100} =66
\frac{\frac{\sqrt{3} }{2}N_{ 1} }{N_{2} } = 57.16
Furnace currents are
I_{a} =\frac{400\times 1000}{100\times 0.707} = 5658 A; \phi _{a} = 45° lagging
I_{b} =\frac{800\times 1000}{100\times 1}=8000 A; \phi _{b} = 0°
Furnace voltages and currents are drawn in the phasor diagram of Fig. 3.69(a).
On the 3-phase side
I_{A} =\frac{5658}{57.16} =99 A
I_{BC} =\frac{8000}{66} =121.2 A
From the phasor diagram of Fig. 3.69(b)
\bar{I_{B} } =121.2-49.5\left(0.707 + j 0.707\right) =86.2 – j 35
or \bar{I_{B} } =93 A
\bar{I_{C} } =121.2 + 49.5 \left(0.707 – j 0.707\right) = 156.2 – j 35
or I_{C} =160 A