When neutral disconnected Lamp are in series connection,
So, for series connection current is same for both Lamp,
Resistance of Lamp (L1) = (230)²/P
= (230)²/100
= 529 ohm
Resistance of lamp (L2) = (230)²/150
= 352.67 ohm
So, current through the lamp = E/(R1 + R2)
= 400/(529 + 352.67)
= 0.454 A
Voltage across lamp (L1) = I*R1
= 0.454*230
= 240 V
Related Question
In FDLF method,
B' corresponds to susceptance of unknown of PQ and PV bus
Order of B' matrix is = (n -1)*(n - 1)
Where n is total no. of buses.
Order of B'' = (n - m - 1)*(n - m - 1)
Conductor with 75 °C temperature rating must have ampacity from the 60°C column or the 75°C column.
It is permissible to use the 75°C rating if the installed conductor is rated at least 75 °C.
Because all of the connection points in this example have at least a 75°C rating, the conductor’s ampacity can be based on the 75°C column