double supermesh problem

Discussion in 'Homework Help' started by hunterage2000, Dec 31, 2010.

1. hunterage2000 Thread Starter Active Member

May 2, 2010
393
0
Im having problems figuring out the equations for the attached supermesh problem I've not seen many examples of these type of problems so not sure what I'm doing.

What I tried to do was:

supermesh 1

(R1+R2+R3)m1 - R3m3 = Vs1 - Vs2 - R2Is1 - R3Is1

supermesh 2

R3m2 + (R3+R4+R5)m3 = Vs2 + Vs3 + R4Is2 + R5Is2

Can someone tell me whats wrong?

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2. kira12a8 New Member

Dec 31, 2010
2
1
With mess current method, you can ONLY use it with independant loops, you can't use it with loops that contain another loop inside it. Wish you good luck man

3. hunterage2000 Thread Starter Active Member

May 2, 2010
393
0
So you would reduce the circuit so their is just one superloop at the right hand side. Thats what I've done.

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I've attached a solution using mesh analysis. There are many ways to do this. I haven't specifically used a supernode technique.

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5. The Electrician AAC Fanatic!

Oct 9, 2007
2,255
311
Using the designations for the mesh currents and the node voltages given by t_n_k in his attachment, we can set up a matrix formulation for the problem. There are two supermeshes, a left one and a right one. With currents in mA and resistances in k ohms, we have:

$\left[ \begin{array}{4}-1 & 1 & 0 & 0 \\5 & 15 & -10 & 0 \\0 & 0 & -1 & 1 \\0 & -10 & 10 & 55 \end{array}\right]\left[ \begin{array}{1}I_A\\I_B\\I_C\\I_D \end{array}\right]=\left[ \begin{array}{1}5\\-5\\\frac{16}{10}\\17\end{array}\right]$

Rows 1 and 3 are the constraint equations, and rows 2 and 4 are the supermesh equations.

Solving gives:

$I_A = \frac{-541}{120}$
$I_B = \frac{59}{120}$
$I_C = \frac{-61}{60}$
$I_D = \frac{7}{12}$

You get all 4 mesh currents at once.

6. hgmjr Moderator

Jan 28, 2005
9,030
214
Using Millman's Theorem allows one to form the formula for the voltage across RL by inspection.

$V_{R_L} =\ \left{ \left[ \Large \frac{\frac{(V_1+R_1I_1)}{R_1+R_2}\ +\ \frac{V_2}{R_3}\ +\ \frac{(I_2(R_4+R_L)\ -\ V_3)}{(R_4+R_L)}}{\frac{1}{R_1+R_2}\ +\ \frac{1}{R_3}\ +\ \frac{1}{R_4+R_L}}\right]\ +\ V_3\ \right}\ \Large *\ \frac{R_L}{(R_4+R_L)}$

I plugged in the circuit values and I came up with the same value for the voltage across RL as t_n_k.

hgmjr