Transformation ratio
Per phase
Secondary phase current
Full-load condition
Full load total Cu loss,
= 4,490 W;
Iron loss = 3,050 W
Total full-load losses
= 4,490 +3,050 = 7,540 W;
Output at unity p.f. = 500 kW
∴ F.L. efficiency = 0.9854 or 98.54%;
Output at 0.8 p.f.= 400 kW
∴ Efficiency = 0.982 or 98.2%
Half-load condition
Output at unity p.f.= 250 kW
Cu losses = 1,222 W
Total losses = 3,050 + 1,122 = 4,172 W
∴ = 0.9835 = 98.35%
Output at 0.8 p.f. = 200 kW
= 0.98 or 98%
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Maximum Power Transfer Theorem :
In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a power source with internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.
Let Impedance of,
Xc near the terminal X is Z1=-j10
10 ohm resistance, Z2=10
3 ohm resistance, Z3=3
XL, Z4 =j14
Xc near to source, Z5=-j10
Now,
Zth=[Z5 || (Z3 +Z4) ] + Z2 +Z1
= [ -j10 || (3+j4)] + 100- j10
= 112 - j36
For t<0, the switch is closed.
Capacitor acts as open to DC.
Voltage across the capacitor,
v(0-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0-)= v(0) = 8VAt t>0, The capacitor is discharging.Rth = (12||4) = 3 ohmC= 1/6 FTime constant, Π = Rth×C = 0.5 So, v(t) = v(0)e-t/Π = 8e-t/0.5 =8e-2t V (Ans.)
The initial energy stored in the Capacitor,
W=(1/2)CVo2
= 0.5×0.167×82
= 5.333J (Ans)
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