I often give my power quality (PQ) 101 course to new engineers or salespeople. This introduction starts with the basics: an overview of PQ phenomena based on IEEE 1159 Table 4.2 followed by Ohm’s and Kirchhoff’s laws and how they apply to power quality. I also still receive data or comments about data from PQ monitors, which would appear to disprove or invalidate these long-standing laws. After further investigation, usually a logical explanation shows that something was connected incorrectly or facts were misinterpreted. Therefore, you must fundamentally understand Ohm’s and Kirchhoff’s laws for virtually all PQ issues.


Ohm’s Law equates voltage to the product of current and impedance. Voltage and current are the most common parameters measured by power quality monitors, along with all of the mathematically derived parameters from them, including power, power factor, harmonics and interharmonics, flicker, unbalance, and frequency. 


Impedance is often an unknown quantity to many users. If the system and load impedances were simple resistors, perhaps it would be more actively discussed. But nearly all electrical systems comprise resistance, capacitance and inductance. In the not-too-distant past, most loads were mainly inductive and resistive, such as electric induction motors. With adjustable speed drives, information technology equipment, and nonincandescent lighting making up the majority of the load on electrical systems, calculating the impedance has become a nonstationary, time-varying, complex mathematical process. But voltage (V) still equals current (I) × impedance (Z).


Kirchhoff’s Law has one method based on voltage and another based on current. In the example above, the node at point A illustrates the first law in which the sum of the currents at a node must equal zero. Stated another way, the current entering the node equals the current leaving it. With an alternating current (AC) system, entering and leaving are a relative term because it reverses each half- cycle of the sinusoidal wave. So we pick one direction as a convention, which is generally that currents go from the “–” to “+” labeled terminals of a source, and enter the “+” and exit the “–” labeled terminals of a load. One key is being consistent with the labeling convention throughout the analysis and letting the mathematics work it out for us. In this example, current I1 enters the node from the source or generator V0 after passing through the impedance Z1. Two currents (I3, I2) leave the node to flow through the loads (Z3, Z2+Z4), respectively. Hence I1= I2+ I3 at both node A and node B, though entering/leaving reverse sides.


The voltage method has a similar zero-sum rule. The voltages around a closed loop, such as the loops labeled 1 or 2 in the example, will equal zero, with the same polarity naming convention applied. V0 equals V1 + V3, and V3 equals V2 + V4. If we take the outer loop, V0 equals V1 + V2 + V4, which we also could have achieved by applying the substitution theorem on V3. 


In the real world, this means the voltage produced by the source (V0) will equal the sum of the voltage drops across all of the loads on the source, including the system impedances (Z1, Z2), which include the wiring, switches, protection devices, etc., and the loads (Z3, Z4). It is immediately obvious that, the farther from the source, the less voltage is available for a load because there are more losses in the system impedances.


If we further simplify the example to include only one loop, the math behind why voltage sags occur can be more easily explained. Let’s say V0 = 110, Z1 = 1, and Z3 = 10 when initially turned on and 100 ohms when running steady state. This is similar to how a motor load performs. If we put a switch between node A and Z3 and open it, there is 110V at V0 and no voltage V1 across Z1 because no current is flowing through it (remember Ohm’s Law); hence, there is 110V at node A. When the switch is closed and load Z3 is initially energized, its impedance is 10 ohms, which is added to the 1-ohm source impedance Z1 to equal 11 ohms. Dividing 110 by 11 equals 10 amperes, and 10 amperes through Z1 is a 10V drop, leaving only 100V for load Z3. Once steady-state conditions are obtained, Z3 has changed to 100 ohms. 


Now 110V divided by 101 (100 + 1) ohms changes the current to 1.09 amperes or approximately 1.1 amperes. With 1.1 amperes flowing through the 1-ohm source impedance equating to a 1.1V drop, this leaves plenty of voltage (108.9V) for the load. This is just like sags generated by loads with large in-rush currents switching on or momentary arcing faults on the power system.


Adding inductance and capacitance to the impedances makes the ohms value vary over frequency, such as when harmonics are present. They also cause a change in the currents and voltage phase relationship, making power no longer simply voltage multiplied by current. But despite the more complex math, the same two laws apply to harmonics and interharmonics, transients, flicker, swells and unbalance, just as it did in the sag example. 


Learn the laws, and never forget them. They will always apply.