ideal transform why there is negative in I2 mash 1 and I1 mash 2?
Thread Starter full Joined May 3, 2014 225 Jan 6, 2015 #1 ideal transform why there is negative in I2 mash 1 and I1 mash 2?
shteii01 Joined Feb 19, 2010 4,644 Jan 6, 2015 #2 If you ASSUME that clockwise direction is positive. Then I2, which is in counter clockwise direction, is negative. 8(I1)+(-4j)[I1-(-I2)]+V1-100=0 8(I1)+(-4j)[I1+I2]+V1=100 8(I1)+(-4j)I1+(-4j)I2+V1=100 (8-4j)I1-4j(I2)+V1=100
If you ASSUME that clockwise direction is positive. Then I2, which is in counter clockwise direction, is negative. 8(I1)+(-4j)[I1-(-I2)]+V1-100=0 8(I1)+(-4j)[I1+I2]+V1=100 8(I1)+(-4j)I1+(-4j)I2+V1=100 (8-4j)I1-4j(I2)+V1=100
WBahn Joined Mar 31, 2012 29,979 Jan 6, 2015 #3 Uhm... because the impedance of a capacitor is negative? What is your reasoning for why you think it should be addition and not subtraction -- there are a couple of possible reasons for the misunderstanding.
Uhm... because the impedance of a capacitor is negative? What is your reasoning for why you think it should be addition and not subtraction -- there are a couple of possible reasons for the misunderstanding.