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Below is a circuit schematic of sources and resistors. $$I_s=2A, V_s=20V, R_1=10Ω, R_2=6Ω, R_3=80Ω, R_4=15Ω$$ enter image description here

Question 1: Calculate the current i1 flowing through resistor R1.

Question 2: For the same circuit, calculate the current i3 flowing through resistor R3.

Question 3: For the same circuit, calculate the current iVs which is flowing through the voltage source from node c to the ground.

I decided to use mesh analysis in this problem:

enter image description here

I thought the currents for each of the following questions are the following:

Question 1: The current flowing through R1 is just i1.

Question 2: Current through R3 is i1+i3.

Question 3: Current through the voltage source is i2 + i3.

Assembling my equations I get the following matrix:

\begin{pmatrix} R_1 + R_2 + R_3 & -R_2 & -R_3 & 0 \\ 0 & i_s & 0 & 0 \\ -R_3 & 0 & R_3+R_4 & V_s \end{pmatrix}.

When solving the the matrix, however, I do not get the right answers:

enter image description here

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  • \$\begingroup\$ Current flowing through R3 is not i1+i3. Current flowing through R2 is not i2 + i3. Edit: looking at the image you posted, you did not make the mistake I commented on above, just a typo on the post. \$\endgroup\$
    – PacEE
    Jan 20 '15 at 4:05
  • \$\begingroup\$ So what is the current through R2 and R3? \$\endgroup\$ Jan 20 '15 at 9:29
  • \$\begingroup\$ @user1527227: They are \$i_2\$ and \$i_3\$ with values given in my answer below. \$\endgroup\$
    – Matt L.
    Jan 20 '15 at 10:22
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I get the following equations

$$\begin{align} i_1R_1 - i_2R_2-i_3R_3&=0\\ i_1+i_2&=I_s\\ i_3R_3+i_4R_4&=V_s\\ i_1+i_3-i_4&=0 \end{align}$$

With the given values this results in

$$i_1=1.007353\;A\\ i_2=0.992647\;A\\ i_3=0.051471\;A\\ i_4=1.058824\;A$$

And the current through the voltage source is simply \$i_2-i_3\$.

EDIT:

Your equations on paper are correct, but you made a mistake setting up your matrix. It should be

\begin{pmatrix} R_1 + R_2 + R_3 & -R_2 & -R_3 & 0 \\ 0 & 1 & 0 & I_s \\ -R_3 & 0 & R_3+R_4 & V_s \end{pmatrix}

Solving this system gives the mesh currents:

$$i_{m1}=1.0074\;A\\ i_{m2}=2.0000\;A\\ i_{m3}=1.0588\;A$$

Note that the mesh currents are generally not equal to the currents in the diagram. The actual currents are given by

$$\begin{align}i_1&=i_{m1}\\ i_2&=i_{m2}-i_{m1}\\ i_3&=i_{m3}-i_{m1}\\ i_4&=i_{m3}\end{align}$$

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  • \$\begingroup\$ Thanks Matt. It seems that your solution is correct. Do you know why my method does not work? Could you please explain what I am doing incorrectly? \$\endgroup\$ Jan 20 '15 at 18:50
  • \$\begingroup\$ @user1527227: I found the mistake in your method, check my updated answer. \$\endgroup\$
    – Matt L.
    Jan 20 '15 at 19:44
  • \$\begingroup\$ Wow thanks so much. So clear. Last question if you don't mind: What exactly is the mesh current? You can see from the diagram in my handwriting that the mesh currents around the loop are not value. In other words, the current at different points of the loop are vary. So does the mesh current have a physical meaning? \$\endgroup\$ Jan 21 '15 at 3:31
  • \$\begingroup\$ @user1527227: In general mesh currents do not correspond to any physical current, but you can easily compute the actual branch currents from a superposition of mesh currents. \$\endgroup\$
    – Matt L.
    Jan 21 '15 at 6:25

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