#### LUQMAN

Joined May 27, 2009
1
Hi all, ive just joined this site.

0.1 ohms 0.059 ohms 0.199 ohms
240v A__________B_____________C____________D

10+30+50 10+30 10

50A 30A 10A

HEY I WAS TRYING TO DRAW A TABLE , BUT NOT SURE HOW TO, USING THIS SITE. lol

please can someone help me i need calculate the following,

• the power lost in the cable
• the power developed bu each load
• the efficency of the system
this is related to my electrical conductors assignment.

Thankyou.

#### studiot

Joined Nov 9, 2007
4,998
Well you have the resistances and currents, so you can use OHMS law to calculate the voltage drop on each section.
Successively subtract these from 240 to get the load voltages at B, C and D.

You have a choice of power formulae for the cable power loss. Calculate each section individually and add them together.

Since you don't specifically know the load resistances you only have one formula for the load powers P=IV. Again calculate each individually.

Efficiency = useful power out / power in x100%

So the useful power out is the load power and the power in is the sum of the load power and the power lost in the cable.

Post again with your efforts if you need more help.

Yes tables are difficult in this forum.