A 12,5kW single-phase inductive load operation at a power factor of 0,75 lagging is connected to a 240V supply. Each conductor has a resistance of 0,1ohm. Calculate
a) phase angle
b) current drawn by the load
c) voltage drop better supply and load
d) voltage across the load
e) is the cable suitable for this particular application? Explain
a) Phase angle:
To find the phase angle, we can use the formula:
cos(θ) = power factor
θ = arccos(0.75)
θ ≈ 41.41°
b) Current drawn by the load:
To find the current, we can use the formula:
power (P) = voltage (V) * current (I)
12.5kW = 240V * I
I ≈ 52.08A
c) Voltage drop between supply and load:
Voltage drop (Vd) = current (I) * resistance (R)
Vd = 52.08A * (0.1ohm + 0.1ohm) = 10.416V
d) Voltage across the load:
Voltage across the load (Vload) = supply voltage (V) - voltage drop (Vd)
Vload = 240V - 10.416V = 229.584V
e) Is the cable suitable for this particular application?
To determine if the cable is suitable, we need to consider the voltage drop. The voltage drop between the supply and the load is 10.416V. This represents approximately 4.34% of the supply voltage (10.416V / 240V * 100%). In general, a voltage drop of less than 5% is considered acceptable. Therefore, the cable is suitable for this particular application.