# Finding Mesh Currents

#### Navito

Joined Sep 12, 2012
2
Hello everyone, Im new to this forum. I hope this post is not too elementary.
However, how could you find the requested current in this system?
Any feedback is welcome
Looking forward to hear from you! http://www.flickr.com/photos/86884898@N07/7956178742/in/photostream

#### Dodgydave

Joined Jun 22, 2012
8,382
I3= 4mA
so 4ma *1k= 4volts dropped across the first 1K resistor leaving 2Volts across the 3 parallel resistors.

6V-4v= 2Volts across the 3 parallel resistors( 2k/2k/1k) = 2v/0.5k = 4ma

I1=2mA, so 4mA-I1=2mA, so I0+I2= 2mA

as they are both 2K each the current is split equally

1mA in each, so I0=1mA and I2=1mA

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• lizard

#### lizard

Joined Sep 10, 2012
2
SO simply explained..................by senior.

• Dodgydave

#### Navito

Joined Sep 12, 2012
2
Thanks Dodgydave, great explanation!!

#### bretm

Joined Feb 6, 2012
152
That doesn't look right. The 4mA is a superposition of i3 and i2. So the top 1k doesn't have 4mA and doesn't drop 4V.

• ErnieM

#### bretm

Joined Feb 6, 2012
152
Let's call the top resistor R1 and it drops V1, the left-most resistor R2 and it drops V2, the bottom-center resistor R3 and it drops V3, and the bottom-right resistor is R4 and drops V4.

From inspection we know V1 / 1000ohm = i3

We know that i1 = 2mA, so we can see V2 / 1000ohm = 2mA - i3.

On the right-hand side we can see V3 / 2000ohm = -2mA - i3

From the center node we can see V4 / 2000ohm = 4mA + i3

So all four voltages are in terms of i3. What value of i3 is correct? We know that V1 - V2 - V3 + V4 have to add up to -6V (look for the path). The only value that makes this work is i3 = -(2/3) mA.

Since 4mA = i2 - i3, we have i2 = (10/3) mA.

That means i0 = -(4/3) mA.

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#### Dodgydave

Joined Jun 22, 2012
8,382
Let's call the top resistor R1 and it drops V1, the left-most resistor R2 and it drops V2, the bottom-center resistor R3 and it drops V3, and the bottom-right resistor is R4 and drops V4.

From inspection we know V1 / 1000ohm = i3

We know that i1 = 2mA, so we can see V2 / 1000ohm = 2mA - i3.

On the right-hand side we can see V3 / 2000ohm = -2mA - i3

From the center node we can see V4 / 2000ohm = 4mA + i3

So all four voltages are in terms of i3. What value of i3 is correct? We know that V1 - V2 - V3 + V4 have to add up to -6V (look for the path). The only value that makes this work is i3 = -(2/3) mA.

Since 4mA = i2 - i3, we have i2 = (10/3) mA.

That means i0 = -(4/3) mA.
There must be MINIMUM of 4mA through the 1K resisitor at the top, as that is the highest current flowing in the circuit! ( the circuit wont work as its just theory)

#### t_n_k

Joined Mar 6, 2009
5,448
Hi there - bretm is correct. Io is -4/3 mA.

#### ErnieM

Joined Apr 24, 2011
7,993
There must be MINIMUM of 4mA through the 1K resisitor at the top, as that is the highest current flowing in the circuit! ( the circuit wont work as its just theory)
It sounds like you have made a conclusion before you made an analysis. I've run the numbers and arrived at the same answer as bretm did (he beat me to posting it so I did not bother).