Solution:
a) Given, V = 230V
P = 750W
T = 30/60 = 0.5 hours
i) Let the maximum current be I
As we know that,
P = VI
Evaluating value of current,
750 = 230I
I = 3.26A
ii) Electric energy consumed, E = PT
E = 0.75kW×0.5 hours
E = 0.375 kWh
Therefore, the number of units used in 30 minutes = 0.375
b) For example, the electric iron will require a fuse with a current rating of 5A because the maximum current for the specified iron is 3.26A.