Winner of Electrical Circuits Contest #1

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Electrical Circuits, Electrical Circuits Contests

Problem

Find $I_x$ and $I_y$ :

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.

Comments

8 responses to “Winner of Electrical Circuits Contest #1”

November 2, 2013

Sakhawat Asad

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

Reply
1. November 2, 2013
  
  Yaz
  
  Hi Sakhawat,
  
  It is over for now. Hopefully, we will have a new contest soon.
  
  Reply
  1. October 30, 2015
    
    kasyap
    
    When is the next contest held sir (Date and year)?
    
    Reply
August 12, 2014

Dharmendra Sharma

How can Ix be -ve since its direction is correct. I has to be +ve….

Reply
November 6, 2014

John

If Iy is -1A then Ix+Iy-2A=0 shouldn’t give Ix=3A?

Reply
1. December 14, 2014
  
  Yaz
  
  A negative sign was missing from the KCL. It is corrected now.
  
  Reply
June 16, 2015

sowji

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

Reply
1. October 31, 2015
  
  kasyap
  
  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.
  
  Reply