check plugs and wires first. Then fuel filter, ait, then tps then map then cranshaft sensor, The order is from cheap to expensive, You can check the sensor voltage just get a manual. Last but not least check the o2 then the cat. If non of these are at fault its more that likley a mechanical, check compression for head gasket then the timing check. Start with having it scanned then this is from low to high in price. I was lucky i checked all this stuff then found the wires were bad, 39 bucks and she ran fine, Had i followed this list i would have found it before checking all the other stuff. But i know they are good now. Good luck let others know this list. Cheap to expensive possible problems with deawoo loss of power problems.
slow acceleration and loss of power is usually down to airflow meter not performing correctly,tests need to be carried out with scan tool if one available
An inductor is a device which stores energy as a magnetic energy.... Ideal inductor have no resistance.....so there is no power loss.... power loss = (I*I)*R
the catalyser can couse loss of power problems if its deteriorated
Phase loss is the loss of power to a specific area of the circuit. Phase loss can result from exposed wires or damaged wires or even downed power lines.
Gravity! We fight that loss of power by using more torque.
P = I2R Where: P = power loss I = current R = resistance
I have the CD, but how do I reset my eclipse 5342 CD car stereo after power loss?
voltage drop is the loss or drop that occured across the element so that voltage gets down and current increases across the element and power loss is like i2r loss and like wastage of power without consuming
loss of compression, loss of power, oil is "milky"
Massive power loss in a turbo vehicle could be due too boost leak for the turbo.
The Insertion Loss of a line is the ratio of the power received at the end of the line to the power transmitted into the line.
Power = voltage times current, and the power loss is the loss in the line, I^2 * R. At 11,000 volts, the current will be (11,000 / 415 = ) 3.77% of what it is at 415 volts. So the power loss in the line at 11,000 volts will be (3.77% ^2 = ) .14% of what it is at 415 volts.