Strictly speaking, the circuit has 5 nodes, at points labelled a, b, c, d and f. If you use the modified nodal analysis to solve the circuit, you'll apply KCL at all nodes but one (usualy the reference one) to end up with a system of lineal independent equations. So, you'll apply KCL at nodes a, b, c and d to determine the (initially) unknown voltages at these nodes.
However, this is modified by the presence of voltage sources connected to ground at nodes a and c. Due to the presence of these voltage sources, the voltages at nodes a and c are not unknowns, so you have only 2 "real" unknowns (voltages at b and d), and you have to write only 2 KCL. If you have to choose, you would like to write KCL at b and d, because writing the KCL at a and c involves the currents flowing through the voltage sources, and you don't know how to express these currents in terms of the node voltages, so you avoid writing KCL equations at nodes having voltage sources connected to ground.
So finally you have to write the KCL at b and d to solve the circuit, and you can also get rid of the KCL at d if you add the impedances of the resistor and the capacitor together to have 10-10j Ohm.
So your first equation (KCL) is ok, and all you have to do from that is to express the currents in terms of node voltages and impedances:
$$I_1+I_2−I_3−I_4=0$$
$$I_1 = \frac{E_1-V_b}{-10j}$$
$$I_2 = \frac{E_2-V_b}{+20j}$$
$$I_3 = \frac{V_b}{10-10j}$$
$$I_4 = \frac{V_b}{10}$$
And if you substitute these currents in the first KCL and solve for Vb you obtain:
$$V_b = \frac{1-j}{1-3j} * (2E_1-E_2) = \frac{1}{2-j} * (2E_1-E_2)$$
Knowing Vb (and E1 and E2) allows you to easily determine all other circuit variable.
None of the given answers really address the question. The mesh analysis approach allows us to reduce the number of degrees of freedom significantly. In the example, mesh analysis provides 3 variables while there are 10 total edges in the circuit. So clearly, we could get and solve 10 equations and 10 variables, but mesh analysis tells us we only need to solve 3.
Two facts are true to allow this reduction.
- Using the mesh currents approach will satisfy the node and loop rules
- A set of currents and voltages that obey both rules is unique.
The first is easy to see, but the second explains why mesh analysis works. Since the mesh currents satisfy the node and loop rules, it must be the only solution.
It is not straightforward to see why statement 2 must be true. The proof by contradiction assumes that two such current-voltage arrangements exist satisfying the node and loop rules. It then shows that you can find a loop where the sum of voltage drops is non-zero.
https://math.stackexchange.com/questions/1742680/unique-solution-for-circuits-in-linear-algebra
Best Answer
The first two equations you wrote are correct. You made mistake while re-arranging them. The equations should be:
$$5 + V_i = 2k\times I_1 - 1k\times I_2\tag1$$ $$5 = 1k\times I_1 - 2k\times I_2 \tag2$$
Solving which you will get the value for \$I_2\$. $$I_2 = \frac{-5+V_i}{3k}$$
And then, $$V_o = I_2 \times 1k = (V_i-5)/3$$
It was more easy if nodal analysis or superposition theorem was used.