Given,
S = 25 kVA, V1/V2 = 2200/220 V, R1 = 1 Ω, R2
= 0.01 Ω,p.f = 0.8, Pcore = 0.8Pcu, FL
Refer secondary resistance to primary,
R2 = = 1 Ω
Full load current,
Copper loss,
Iron loss,
Efficiency,
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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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