Can someone explain to me how to do this question please. thanks
K Thread Starter kd3041 Joined Feb 2, 2011 7 May 4, 2011 #1 Can someone explain to me how to do this question please. thanks Attachments Untitled.png 30.3 KB Views: 32
R!f@@ Joined Apr 2, 2009 9,918 May 4, 2011 #2 A1 = A2+A3, A3 = A1-A2. All meters give RMS readings. U do the math
K Thread Starter kd3041 Joined Feb 2, 2011 7 May 4, 2011 #3 really? thats too simple. i thought the same thing but i wasnt sure
Jony130 Joined Feb 17, 2009 5,488 May 4, 2011 #4 Look at Phasor diagram for parallel circuit http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE6203
Look at Phasor diagram for parallel circuit http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE6203
R!f@@ Joined Apr 2, 2009 9,918 May 4, 2011 #5 All true Jony, but OP's diagram gives nothing to compare phase difference or anything. Just simple math and RMS confusing thing to n00bs {ed} If there were an Inductor then that would be another story.
All true Jony, but OP's diagram gives nothing to compare phase difference or anything. Just simple math and RMS confusing thing to n00bs {ed} If there were an Inductor then that would be another story.
Jony130 Joined Feb 17, 2009 5,488 May 4, 2011 #6 Well for AC signals and RC/LR components the mother nature do not add "linear" but in geometric mean. \( I_{total} = \sqr{(I_{R}^2 + I_{C}^2) \) So the answer is \( I_{R}^2 = \sqr{(I_{total}^2 - I_{C}^2)} = \sqr{7} = 2.64575131A \) Last edited: May 4, 2011
Well for AC signals and RC/LR components the mother nature do not add "linear" but in geometric mean. \( I_{total} = \sqr{(I_{R}^2 + I_{C}^2) \) So the answer is \( I_{R}^2 = \sqr{(I_{total}^2 - I_{C}^2)} = \sqr{7} = 2.64575131A \)
