June 08 paper 2 question 6(b)
6 (b) 1st is open so power supply so that is zero
2nd from the first the power is supplied so it will move to s
2 but s
3 is closed so it will not move from s
3 so only of s
2 which is 1.5kW so for second it is 1.5kW
3rd the power is supplied from s
2 and s
3 so both are added 1.5kW+1.5kW=3kW
4th - Current flows through A and B (not the part of the junction with S
2 as it has infinite resistance)
R = 38.4ohms, V one heater receives is 120V , P = V
2/R = 120
2/38.4 = 375W
Thus, for two heaters = 375*2 = 750W or 0.75kW
5th - Current flows through all heaters. Power in A and B as calculated previously = 0.75kW
Power in C = V
2/R = 240
2/38.4 = 1.5kW Total - 1.5 + 0.75 = 2.25kW