Winner of Electrical Circuits Contest #1

Problem

Find I_x and I_y :
Electrical Circuit Contest #1

Solution

Three resistors are in series and their equivalent, 6\Omega, is parallel with the voltage source. So, according to the Ohm’s law: I_y=-\frac{6V}{6 \Omega}=-1 A. The negative sign comes from the direction I_y.
Applying KCL at the bottom node:
-(-2A)+I_x+I_y=0 \rightarrow I_x=-1 A.
The lucky winner of the Electrical Circuits Contest #1 is Kunal Marwaha from UC Berkeley. I would like to say thank you to all participants and I am thinking of holding contest #2 soon. Kunal, congratulations and soon you will receive the prize by Paypal.

Published by Yaz

Hi! Yaz is here. I am passionate about learning and teaching. I try to explain every detail simultaneously with examples to ensure that students will remember them later too.

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8 Comments

  1. sir , can u give some information about electrical circuit contest…. . how can i participate?

  2. plz say clearly about this problm applying about kcl
    suppose applying kcl to botem node how the iy in eqution

    1. 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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