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I feel really dumb asking this, but why is a resistor always a drop in loop analysis?

Loop

For instance, in the picture the KVL #2 equations are the correct ones, but why? Shouldn't the resistors keep the sign convention you assign to it with the first equation?

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Resistors are always a drop because they remove energy from the circuit in the form of heat. You can assign any direction you want to start your analysis. When your are done, the result will tell the actual direction. If you end up with Ix = -3 then it means the current is flowing in the opposite direction from your assumption.

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  • \$\begingroup\$ Thank you, I assume capacitor and inductor impedance is the same? \$\endgroup\$ – nbstrong Nov 3 '16 at 16:03
  • \$\begingroup\$ Almost the same. From an analysis point of view you follow the same procedure, with jwL and 1/jwC. These devices don't dissipate energy they store it and return it later. I didn't really look at your equations but the way I would write the second loop is (-3-2)I2+2I1. The -2 and -3 for I2 are the drops in the direction I'm going. The +2I1 is because I1 was was assigned the other direction through the resistor. \$\endgroup\$ – owg60 Nov 3 '16 at 16:27
  • \$\begingroup\$ I guess I mean in that a cap and ind are also always a drop? \$\endgroup\$ – nbstrong Nov 3 '16 at 16:30

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