Kirchhoff’s Current Law – How Does Kirchhoff’s Current Law Work for Sinusoidal Voltage Sources?

Recall that, for node voltage analysis, a floating voltage source (a voltage source that does not connect to the GND node) poses a problem since you cannot write an equation relating the current through to the voltage across.

What you must do then is enclose the floating voltage source in a supernode, which reduces the number of KCL equations by one, and add the equation relating the voltage difference between the nodes the voltage source is connected to.

Now, the dual of node voltage analysis is mesh current analysis and here we have the dual problem when we have a current source common to two meshes - we can't write an equation relating the current through to the voltage across a current source.

What must be done then is to form a supermesh which reduces the number of KVL equations by one and add the equation relating the difference of the mesh currents to the common current source.

So, write KVL counter-clockwise around the supermesh consisting of the two voltage sources and the two resistors

$$V_1 = I_aR_1 + V_2 + (I_b - I_c)R_2$$

You have, by inspection (no KVL required for this mesh - this is dual to no KCL required for the node connected to a non-floating voltage source)

$$I_c = -1.25A $$

You need one more equation which is the equation relating to difference of the two mesh currents with the common current source.

$$3A = I_a - I_b $$

Now, you have 3 independent equations and 3 unknowns.

Here, let me set up the equations for you:

^{simulate this circuit – Schematic created using CircuitLab}

You have five unknowns — three currents and two voltages. We'll need five equations.

KCL tells us that the sum of currents into any node must be zero:

$$I_{L1} + I1 - I_{L2} = 0$$

$$I_{L2} - I2 - I_{L3} = 0$$

$$-I_{L1} - I1 + I2 + I_{L3} = 0$$

Note that the third equation tells us nothing new; it is simply a combination of the first two.

KVL tells us that the sum of voltages around a loop must be zero, which is really just another way of saying that the voltage across I1 must equal the voltage across R1, etc. We simply apply Ohm's Law to the resistors:

$$V1 = -I_{L1} \cdot R1$$

$$V1 - V2 = I_{L2} \cdot R2$$

$$V2 = I_{L3} \cdot R3$$

Now we have five independent equations that we can solve for our five unknowns.

As Alfred Centauri says, the principle of superposition makes this problem almost trivial to solve, but you may not have been exposed to that yet.