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.
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).
You need to remember the identity expressed by equation (3), since it is used often. Each voltage can be expanded and the sum of four voltages is expressed by equation (4).
Since the sinusoidal functions are not zero, we get equation (5), where the coefficients of sinusoidal functions are equated to zero. It can be seen that the sum of cosine terms is zero and that the sum of sine terms is zero.
Equation (5) can be represented, as shown by equation (7). Equation (7) can be represented by equation (8), where Euler's identity, defined by equation (6), has been used to get each of the four terms.
We can express each voltage by a phasor, as shown by equation (9), and we get that the phasor sum of voltages is zero, as shown by equation (10). When the phasor sum is zero, the sum of real parts of phasors is zero, and the sum of imaginary parts of phasors is zero. What has been proved is general. It can be extended to loops with larger number of elements. The same technique can be used to prove Kirchoff's current law also.
The phasor sum of four voltages is zero, as shown by the sketch in Fig. 2. It can be seen that the sum of real parts of phasors is zero and the sum of imaginary parts of phasors is zero. Kirchoff's laws are basic laws, and all other theorems are derived from these laws.
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.
According to Kirchoff's voltage law, the phasor sum of voltages across elements in a closed loop should be zero, and the sign of voltage in loop equation is dependent on the direction of loop current. The loop current enters impedances Z1 and Z2 through their positive terminal, and hence these voltages have a positive sign in front of them in equation (11). On the other hand, the loop current enters the source through its negative terminal, and hence there is a minus sign in front of the source voltage. Move source voltage, VS , to the other side of equation, and then we get the second line of equation (11).
We can express voltages, V1 and V2, as shown by equation (12). Then the source voltage equals the phasor sum of two voltages, as expressed by equation (13). The ratio of source voltage to current from the source is the equivalent impedance, as seen by the source and it is the complex sum of impedances, Z1 and Z2, as shown by equation (14). An example is presented now.
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.
Voltage division rule can be applied to ac circuits also. Here the division is obtained using the phasor value of impedances. As shown by equations (19) and (20), the voltages across the two impedances can be obtained. They are the same values, as obtained earlier. The result can be verified by using Tellegen's theorem. The phasor diagram, drawn for this circuit is presented next.
The phasor diagram, relating the source voltage and the voltages across the two impedances, is shown in Fig. 5. The source current is the reference phasor, and it is seen that the source voltage lags the source current. The source voltage is the phasor sum of the voltages across the impedances. It can be seen that the process of summation is the same as that used for complex numbers. A complex number is a two-dimensional quantity and can be diagrammatically presented by a sketch with two axes. On the other hand, a phasor exists in time with a specified amplitude and a phase, expressed in relation to the reference phasor. The real part and the imaginary part of a phasor can be displayed by a phasor diagram, as illustrated here.
Matlab can be used to analyze circuits. The script for solving this problem is shown below. If you are familiar with Matlab, you can run the program, by copying the script. It is possible to draw the phasor diagram, as shown.
% Solution to Example 1
Z1=4+3*j;
Z2=6-8*j;
Vs=100;
Zeq=Z1+Z2;
I=Vs/Zeq;
V1=I*Z1;
V2=I*Z2;
% Drawing a phasor diagram
clf;
gcf;
hold on;
axis([-20 120 -70 70]);
compass(Vs);
compass(V1);
compass(V2);
text(105,0,'Vs');
text(real(V1)+5,imag(V1),'V1');
text(real(V2)+5,imag(V2),'V2');
V1
V2
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 I1 is 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 I2 is 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.
For the circuit in Fig. 8, we get the following equations. The source current is the phasor sum of currents through the three impedances in parallel, as shown by equation (37). We can express the admittance as the reciprocal of impedance, and the equivalent admittance seen by the source is the sum of three admittances, as shown by equations (38) and (39). It is seen that the current division is proportionate to admittance, as shown by equation (40).
Another example is presented now. The same problem is also presented as the second interactive example, wherein you can solve the problem given different sets of initial conditions.
SOLUTION:
Angular Frequency is given. Obtain XL1, XL2, and XC.
The circuit in Fig. 2_1 is re-presented in Fig. 2_2. The steps required for getting the solution are as follows.
Steps For Solution
1. Voltage across the capacitor is given. Obtain its current IC.
2. Obtain the impedance due to R and L2 .
3. Find the current IRL from VC and the impedance of R and L2 .
4. Find the source current, IS as the sum of IC and IRL.
5. Obtain the drop VL1 across L1 due IS.
6. Add VL1 to VC and get the source voltage VS.
Alternatively, get the admittance of capacitor and the network of R and L2.
Then get the current through R and L2 from the product of this admittance and the capacitor voltage. Add this current to the capacitor current to get the source current. Then follow steps 5 and 6.
The capacitor voltage is known. Hence the capacitor current and the voltage at node A can be obtained as shown below.
As shown by equation (2.3), get the admittance of the network of R and L2. Equation (2.4) , shows the sum of the admittances due to the capacitor and the network of R and L2. The source current is the product of this admittance and the capacitor voltage, as shown by equation (2.5). Equation (2.6) expresses the source current as a function of time.
Another Approach
From the susceptance of the capacitance and the capacitor voltage, get the capacitor current, as shown by equation (2.7). From the admittance of R and XL2 in series and the capacitor voltage, get the current through the resistor, R, as shown by equation (2.8). The expressions for voltage across the resistor R and the inductor L2 are shown by equations (2.9) and (2.10). The source voltage is obtained by getting the phasor sum of the drop across inductor L1 and the capacitor voltage, as shown by equation(2.11). The result is displayed by equation (2.12).
Equation (2.13) expresses the source voltage as a function of time. Remember that the phasor is represented by the peak value of the signal and the phase angle is with reference to reference phasor. The phasor diagram are shown below.
Matlab Script
% Solution to Worked Example 2:
VC = 200 + 200*j; % Given capacitor Voltage
L1 = 0.003; % L1 in Henry
L2 = 0.004; % L2 in Henry
C = 0.001; % C in Farad
R = 2; % Resistance R in Ohms
omega = 500; % omega in rad/s
XL1 = omega*L1; % Reactance of L1
XL2 = omega*L2; % Reactance of L2
BC = omega*C; % Susceptance of C
YC = j*BC ; % susceptance of C
IC = VC*YC;
YRL = 1.0/(R + j*XL2); % Admittance of R and L2
IRL = VC*YRL; % Current through R and L2
VR = R*IRL; % Voltage across R
VL2 = j*XL2*IRL; % Voltage across L2
IS = IC + IRL; % source current
VL1 = IS*j*XL1; % Drop across L1
VS = VC + VL1; % source Voltage
IC
IRL
IS
VL2
VR
VC
VC
VL1
VS
Output from the script
IC =(-1.0000e+002) +(1.0000e+002i)
IRL =100
IS =0 +(1.0000e+002i)
VL2 =0 +(2.0000e+002i)
VR =200
VC =(2.0000e+002) +(2.0000e+002i)
VC =(2.0000e+002) +(2.0000e+002i)
VL1 =-150
VS =(5.0000e+001) +(2.0000e+002i)
Matlab outputs numbers in standard format. The answers obtained are the same.
This page has shown how the Kirchoff's laws can be applied to ac circuits. These laws are used in the subsequent pages. The next two pages present interactive problems. Pages on power factor improvement, start-delta networks, and application of network theorems to ac circuits are presented subsequently.