1. b) The graph drawn is sinuisoidal with peak clearly shown as 340V and 3800V respectively . Draw 2 waves on the same axes marking the peak values only. Note: marking the rms value is unclear as it is somewhere below the peak.
c) To find power loss , use I2r. Current in the secondary is 0.533A . Power loss= 0.533sqX30=8.53W. Power input is 240X6=1440W. Percentage is 8.53/1440 = 0.59%
d) If transmitted at 6A , the power loss = 6sqX30=1080W. Power input = 1440W .
Percentage = 1080/1440=75%
Note use current formula for power loss in cable is most straight forward.
If you use power = V2/R. The V for cable = 0.533x30=16. Power loss is 16 divide by 30 = 8.53W. Giving you the same answer.
Without stepping up, the V = 6x30 =180V. Power loss is 180sq divide by 30 =1080. Hence the same answer 75%.
2. The peak is 330 V , and after stepping down is peak 81.3V. The rms is 230V and after stepping down is 57.5 V.
aiii) the half rectified wave can be drawn on the positive or negative voltage depending on potential of B wrt A or otherwise. Looking at the circuit, current flow from A to B , so A is always higher than B . Drawing A wrt B is positive, and B wrt A is negative.
bi) To calculate power dissipated across R is to calc a half rectified waveform power. To simplify just calc before rectified which is Vms sq divide by R which is 165W . Since for half rectified is having half the waveform only , hence the power is 82.6W.
biii) The power against time is follow the potential against time. The peak voltage is 81.3 hence peak power is 342W . The voltage is zero for half the cycle hence power is zero for half a cycle. Remember to draw turning point for power curve when touching the x axis. Mark the period 0.02 s regularly.
