Finding Mesh Currents

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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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.
 

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)
 

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).
 
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