KVL and KCL in AC Circuits

Kirchoff's Voltage Law

Kirchoff voltage law states, that the algebraic sum of voltages across elements, present in a closed loop, is zero at any instant. In the case of ac circuits, the phasor sum of voltages across elements in a closed loop is zero. As in the case of dc circuits, a loop current is assigned for each loop, and the sign for voltages should be marked, in accordance with passive sign convention. Figure 1 shows a circuit with a single loop, containing four elements. Loop current, i(t), has been assigned clockwise direction, and the voltages across the elements have been marked, in accordance with passive sign convention.

According to Kirchoff's voltage law, equation (1) can be formed. It is known that the sum is zero.

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Let the voltages be as they are shown in Fig. 2, where the voltages are shown as phasors, and their phasor sum is shown to be zero, by the sketch on the right side. We can formally prove it as follows. Let the four voltages be defined, as shown by equation (2). Each voltage is a sinusoidal voltage, with an amplitude and a phase. We can expand each voltage, using the trigonometric identity shown in equation (3)

Let us apply Kirchoff's voltage law to a circuit. An ac series circuit is shown in Fig. 3. The aim is to find an expression for the impedance seen by the source. Let us go through the steps sequentially. Loop Current for this circuit is marked. Based on the direction of loop current, the loop equation can be formed. This equation is shown next.

Sample Problem

A circuit is presented in Fig. 4. The values of the impedances and the current through the circuit have been specified. The task is to find the source voltage and then draw the phasor diagram that shows the relationship between the source voltage, and the voltage across the two impedances. The solution is as shown below.

We get the impedance seen by the source, as expressed by equation (15). The source current is known, as stated in Fig. 4. Then we can obtain the source voltage, the voltages across the impedances, as shown by equations (16), (17) and (18). We can obtain the voltages across the impedances using the voltage division rule, as shown below.

Kirchoff's Current Law

Let us next consider an ac circuit, with two impedances, connected in parallel. We need to apply Kirchoff's current law to such a circuit. An ac circuit with two impedances is shown in Fig. 6. We can apply Kirchoff's current law to node, A. The aim is to find an expression for the impedance seen by the source and the procedure is illustrated below.

For the circuit in Fig. 6, we can apply Kirchoff's current law to node, A. The equation obtained at node, A, is displayed by equation (21). Given the source voltage, the source current is the sum of currents, as shown by equation (21). The current through each of the two impedances can be obtained, as shown by equation (22). Equation (23) expresses the source current as the sum of currents through the two impedances. As expressed by equation (24), the ratio of the source current to source voltage is equivalent admittance of the circuit. The equivalent admittance is the reciprocal of the equivalent impedance, and is the sum of the reciprocals of the two impedances, as shown by equation (24). The equivalent impedance can be obtained, as shown by equation (25).

Equations (26) and (27) express the currents through the impedances. It can be seen that when two impedances are connected in parallel, the current division is proportionate to the admittance. In this case, the current through the impedance is the product of its admittance and the source voltage. In other words, it can be stated that the source voltage is the ratio of source current to the equivalent admittance of the circuit.

An example is presented now, to illustrate what has been described so far. For the circuit in Fig. 6, let the values of elements be defined, as follows.

For the circuit in Fig. 6, the source voltage and the two impedances are as stated by equation (28). The task is to find the source current, and the currents through the two impedances. Finally, the solution is verified by use of Tellegen's theorem.

The equivalent admittance and the impedance can be obtained, as shown by equation (29). Once the equivalent impedance is known, the source current can be obtained, as shown by equation (30). Then the current through impedance Z₁ is obtained by equation (31), and the current through impedance Z₂ is obtained by equation (32). The solution can be checked, as shown by equation (33).

The phasor diagram relating the source current, and the currents through the two impedances is shown by the sketch in Fig. 7. The source current is the phasor sum of the currents through the two impedances. The angle of source current is negative, and the source current lags the source voltage, which is the reference phasor in this context. The angle of current I₁is more negative than that of the source current, and current I₁ lags the source current. Remember that that the positive direction for angle is counter-clockwise. The angle of current I₂is positive, and it leads the source voltage. Next, verification using the Tellegen's theorem is presented.

Verification using the Tellegen's theorem is carried out as follows. Find the apparent power associated with impedance Z₁. We need to use the factor of half, because both the voltage and the current phasors are qualified by their amplitudes. If the root mean square values are used, then we need not use this factor of half. Find the apparent power associated with impedance Z₂, and the apparent power associated with the source. A negative sign is needed for the apparent power of the source, since its current is leaving its positive terminal, and we need a negative sign in such a case, so that passive sign convention is observed. It can be seen that the phasor sum of apparent powers is zero, as stated by Tellegen's theorem. It is necessary to verify the solution to a problem, to ensure that your solution is correct.