Three resistors are in series and their equivalent, , is parallel with the voltage source. So, according to the Ohm’s law: . The negative sign comes from the direction .
Applying KCL at the bottom node:
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It is over for now. Hopefully, we will have a new contest soon.
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How can Ix be -ve since its direction is correct. I has to be +ve….
If Iy is -1A then Ix+Iy-2A=0 shouldn’t give Ix=3A?
A negative sign was missing from the KCL. It is corrected now.
plz say clearly about this problm applying about kcl
suppose applying kcl to botem node how the iy in eqution
Hey, very simple. From the diagram, Ix+Iy=-2A (Assume the bottom node like you said. Ix and Iy are incoming currents. And outgoing current is -2A. As per KCL, sum of incoming currents equals sum of outgoing currents). Keep this as one equation.
Next to loop 2, apply KVL.
You will get (-2-2-2)ix-6=0. This gives -1A current in clockwise direction. Apply this value in first equation. You will get Iy=-1A.
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