Can we solve this problem using supermesh analysis? I thought you needed to have a common current source between each mesh? Is there an easier solution to this problem?
Electronic – How to solve this problem using supermesh analysis
circuit analysiscurrentmeshvoltage
Related Solutions
The currents split up as shown in the following schematic:
simulate this circuit – Schematic created using CircuitLab
You could easily combine \$R_1\$ and \$R_2\$ into a single effective value and work out the voltage at their shared top node... and from that work out their currents. But you can also just as well see that \$R_1\$ is 3 times less conductive than \$R_2\$, so this means that if one part of the current flows through \$R_1\$ then it must be that three more parts of the current must flow through \$R_2\$. From this argument, you should consider dividing the total current of 12 A into four parts of 3 A each. And thus, \$i_1=3\:\textrm{A}\$ and \$i_2=9\:\textrm{A}\$.
Similar logic then applies to the other two resistors. Here, it's not hard to imagine a total of 7 parts of the -2 A net total current in \$R_3\$ and \$R_4\$, where now three parts go through \$R_3\$ and four parts through \$R_4\$. Each part is \$-\tfrac{2}{7}\:\textrm{A}\$, so this means \$i_3=-\tfrac{6}{7}\:\textrm{A}\$ and \$i_4=-\tfrac{8}{7}\:\textrm{A}\$.
Looking at the bottom node, we know the following must be true:
$$ -8\:\textrm{A} - 2\:\textrm{A} + 3\:\textrm{A}+9\:\textrm{A}+-\tfrac{6}{7}\:\textrm{A}+-\tfrac{8}{7}\:\textrm{A}=0$$
And it is.
simulate this circuit – Schematic created using CircuitLab
Applying KVL on Mesh 1 :
80 = 10i1 + 20(i1–i2) + 30 (i1–i3)
80 = 10i1 + 20i1 - 20i2 + 30i1 - 30i3
80 = 60i1 – 20i2 – 30i3... ... (1)
Applying KVL on the Supermesh :
30 = 40i3 + 30(i3–i1) + 20(i2–i1)
30 = 40i3 + 30i3 – 30i1 + 20i2 - 20i1
30 = 70i3 – 50i1 + 20i2 ... ... (2)
Applying KCL at Node1 :
15i1 = i3 – i2
i3 = 15i1 + i2 ... ... (3)
Solving (1), (2) and (3) :
i1 = 0.583 A
i2 = -6.15 A
i3 = 2.6 A
Result :
v3 = 40 * i3
v3 = 104 V
Best Answer
Try this:
One loop on the left side:
$$ -1V + I_L\:50\Omega + I_L\:1k\Omega = 0$$ (1)
And two loops on the right side \$I_1\$ and \$I_2\$.
And for \$I_2\$ loop we can write mesh equation like this:
$$ I_2\: 5k\Omega + I_2\:100\Omega (I_2 + I_1)50k\Omega = 0$$ (2)
For the \$I_1\$ loop we do not need to write a mesh equation because we have a current source in it hence, the \$I_1\$ mesh current must be equal to VCCS current.
$$I_1 = 40S \times V_P $$
Additional we know that:
$$V_P = I_L \times 1k\Omega $$
And finally, we have:
$$I_1 = 40S \times V_P = 40S \times \:I_L \times 1k\Omega $$
Now we can substitute this into equation 2 thus we end up with this two eqiations:
$$ -1V + I_L\:50\Omega + I_L\:1k\Omega = 0$$ $$I_2\: 5k\Omega + I_2\:100\Omega (I_2 + \left(40S\:I_L\:1k\Omega )\right)50k\Omega = 0$$
And the solution is:
$$I_L = 0.952mA$$ $$I_2 = - 34.569A$$
And from the Ohm's law we have
$$V_O = I_L \times 5k\Omega =- 34.569A \times 5k\Omega = -172.845kV $$