# 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.

## 8 thoughts on “Winner of Electrical Circuits Contest #1”

1. Sakhawat Asad says:

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

1. Hi Sakhawat,

It is over for now. Hopefully, we will have a new contest soon.

1. kasyap says:

When is the next contest held sir (Date and year)?

2. Dharmendra Sharma says:

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

3. John says:

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

1. A negative sign was missing from the KCL. It is corrected now.

4. sowji says:

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

1. kasyap says:

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.