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Bog'liq
Electric Circuit Analysis by K. S. Suresh Kumar

I
I
I
(a)
(b)
r
R
R
R
3
r
3
r
3
r
N
3
I
V
P

V
P

V
P

+

+
+


I
R
I
Y
I
n
I
n
I
B
R
R
R
R
R
3
r
3
r
B
Y
3
r
N
V
P
V
P

V
P
V
P
= (1 + 1

)
+

+
+


β
β
α
α
Fig. 8.1-2 
Fromsingle-phasecircuittothree-phasecircuit
Next, we imagine that the negative polarity terminals of all the three sources are joined together to 
form a common node and that that a single cable connects this node to a similarly formed node on the 
load side as shown in the circuit of Fig. 8.1-2(a). The return currents from the three load branches flow 
through this cable. Then, this cable must have three times the cross-sectional area of the other cables 
in circuit in Fig. 8.1-2(a). Therefore, its resistance will be r. The power loss in four cables put together 
will again be 18r (V
P 
/R)
2
W.
Non-uniform heating will result if the three branches get different powers delivered to them. 
Therefore, we wish to keep the power dissipated in the three resistors equal. Therefore, the three 
sources must have the same rms voltage. But that does not mean that they should be in-phase 
too. We are not particularly worried about the relative phase of currents flowing in the three 
branches of a heater resistance. However, is there any advantage in making the phase angles
non zero?
We assume in circuit of Fig. 8.1-2(b) that the phase of source connected at the terminal identified 
Y is at an angle 
a
with respect to the source connected at terminal marked as R. Similarly, the 
phase of source connected at the terminal identified as B is at an angle 
b
with respect to the source 
connected at terminal marked as R. Moreover, we assume that now we are using an extra-thick 
cable between node N to a similar node on the load side such that it is virtually a resistance-free 
connection. Now, the currents flowing out from the sources have different phase; but same magnitude 
of V
P
/(R

3r) A rms. Note that we use ‘R’ as a terminal marking and ‘R’ as symbol and value for 
a resistance. The cable connecting the node N to a similar node on the load side will have a current

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