simple circuit problem using KVL and KCL

PG1995 · Apr 19, 2011

Hi

Please have a see on this circuit diagram:
https://img192.imageshack.us/img192/7726/meshkvl.jpg

Note: "1", "2", "3" coming after the "I" are subscripts.

I used these equations:
1: I1 = I2 + I3 (Using KCL at junction e)
2: 6 - 4I2 + 4 - 2I2 = 0 (Using KVL around afeba)
3: 3 - 6I3 - 4 - 4I2 = 0 (Using KVL around dcbed)

I2 = 2.8 A

I1 = 4.5 A

I3 = 1.7 A

I don't understand where I'm wrong. Perhaps the error lies somewhere in the approach. Would you please help me with it? Thanks.

MikeMl · Apr 19, 2011

First, lets see what LTSpice says: (note I swapped the position of V2 and R2 to make it easier to visualize. I also labeled the nodes differently than you).

Now, using algebra:

Code:

Σ of currents into node B = 0 

I1 + I2 + I3 = 0

I = V/R


[V(c)-V(b)] + [V(d) -V(b)] + [V(e)-(V(b)]
--------      -----------    -----------          =   0
    2             4              6


6[-6-V(b)] +  3[4-V(b)]  +  2[3-V(b)]
-----------------------------------            = 0
               12

-36 -6V(b) +12 -3V(b) +6 -2V(b) = 0 


-3V(b) -6V(b) -2V(b) = 36 -6 -12

-11V(b) = 18

V(b) = -18/11 = -1.36363


Now that you know V(b), you can solve for the branch currents.

Ratchit · Apr 19, 2011

MikeMl,

I believe you reversed the battery V3. The voltage at node b should be -18/11 = -1.64 .

PG1995,

The node equation is (Vb+6)/2 + (Vb-4)/4 + (Vb-3)/6= 0 ======> Vb=-18/11 = -1.64

Ratch

BrownOut · Apr 19, 2011

PG1995 said:
2: 6 - 4I2 + 4 - 2I2 = 0 (Using KVL around afeba)

Should be 6 - 4I2 + 4 - 2I1 = 0, Yes?

PG1995 said:
3: 3 - 6I3 - 4 - 4I2 = 0

Should be 3 - 6I3 - 4 + 4I2 ( look at the direction of I2 )

PG1995 · Apr 20, 2011

MikeMl, Ratchit, BrownOut, I'm very thankful to all of you. MikeMl, I do have some questions on LTspice which I will ask soon. I have it installed it on my computer but don't know how to use it. I also have Qucs program.

Once again, thank you.

simple circuit problem using KVL and KCL

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simple circuit problem using KVL and KCL

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