$$\underline{\textbf{Power Factor Correction on a Simplified 12 kV Distribution System}}$$
This simplified 12 kV distribution system demonstrates how capacitors can be used to offset inductive loads and improve power factor.
$$\underline{\textbf{Purely Resistive Load}}$$
The branch on the far right with switch (#SW4), containing the 40Ω resistor (#R4), represents a purely resistive load.
With 12 kV RMS applied across the resistor, the expected current is:
$$I = \frac{V}{R} = \frac{12,000\ \mathrm{V}}{40\ \Omega} = 300\ \mathrm{A}$$
This matches the measured branch current (#I_R4) when the switch (#SW4) is closed.
Because the load in this branch is purely resistive, its power factor is 1.0 (100%), meaning the current through (#R4) in orange is perfectly aligned with the applied voltage (#V_1) shown in green. With no other loads connected, the branch current (#I_R4) is identical to the total source current (#I_T). Consequently, the source current and the system voltage (#V_1) in green remain in phase.
$\textit{(Note: When this is the only branch enabled, the branch current trace}$ (#I_R4) $\textit{and the total source current trace}$ (#I_T) $\textit{overlap completely and appear as a single line.)}$
$$\underline{\textbf{Inductive Motor Load}}$$
The middle branch with switch (#SW3) represents a motor containing both resistive and inductive elements.
$\textit{(Note: After closing the motor switch}$ (#SW3}, $\textit{it may take 10 to 20 seconds for the simulation to settle and for the displayed values to stabilize. Don't worry about}$ (#RL3), $\textit{it's needed for the simulation to dampen oscillations when}$ (#SW3) $\textit{is first closed, and can be ignored.)}$
The inductor (#L3) models the motor windings required to create the magnetic field, while the parallel 40Ω resistor (#R3) represents the portion of the motor load that performs useful work.
When this branch is enabled, you might expect the total source current (#I_T) to increase by another 300A because resistor (#R3) is also 40Ω of load in parallel with the source (#G1). Instead, the total source current (#I_T) increases from 300A to approximately 916A.
The additional current is caused by the 50 mH inductor (#L3). Unlike the resistor (#R3), the inductor (#L3) continuously stores energy in its magnetic field and then returns that energy to the source (#G1). This energy transfer requires current to flow, even though that current is not performing any useful work in the motor. Only the current passing through (#R3) is performing work.
Notice that the total source current (#I_T) shown in blue is now out of phase with the system voltage (#V_1) shown in green. A significant portion of the current is circulating between the source and the inductor rather than being converted into useful work in either (#R3) or (#R4).
With this motor load connected, the feeder must supply approximately 916A (#I_T), even though only a portion of that current is associated with real power (#I_R3). The remainder is reactive current (#I_L3) required to sustain the magnetic field in the motor. This reduces the power factor from 1.0 (100%) to approximately 0.656 (65.6%).
$$PF = \frac{I_{\text{real}}}{I_{\text{total}}} = \frac{600\ \mathrm{A}}{914\ \mathrm{A}} = 0.656$$
$$\underline{\textbf{Capacitor Bank Correction}}$$
The branch on the far left (#SW2) represents a capacitor bank (#C2) similar to those commonly installed on distribution system lines.
When the capacitor (#C2) is connected by closing (#SW2), it forms a resonant, or "tank," circuit with the motor inductance. Energy is now exchanged (#Itnk) locally between the capacitor (#C2) and the inductor (#L3), reducing the amount of reactive current that must travel all the way back to the source.
Notice that the current supplied by the source (#I_T) drops from approximately 916 A to about 667 A after the capacitor bank is connected. The motor is still using the same amount of real current (#I_R3), but much of the reactive current (#I_L3) is now being supplied locally by the capacitor (#C2). As a result, the system power factor improves from 0.656 (65.6%) to approximately 0.901 (90.1%), significantly reducing the total current (#I_T) that must be carried by the feeder.
$$PF = \frac{I_{\text{real}}}{I_{\text{total}}} = \frac{600\ \mathrm{A}}{667\ \mathrm{A}} = 0.901$$
Although the motor still requires the same amount of real power, the system now needs far less current from the source feeder (#G1) because much of the reactive power is being supplied locally by the capacitor bank. This is the fundamental purpose of power factor correction on an electrical distribution system.
$$\underline{\textbf{An Important Observation}}$$
Notice that the current circulating within the inductor-capacitor tank circuit (#Itnk) shown in red, is still approximately 916A, which is actually greater than the current being supplied by the source (#I_T).
Capacitors like (#C2) do not eliminate the inductive load current (I_L3). Instead, they provide a local source of reactive power so that the energy exchange between the inductor and capacitor occurs near the load (#Itnk) rather than through the entire distribution system.
This reduces the current carried by upstream conductors (#I_T), freeing feeder capacity, reducing losses, and improving overall system efficiency while allowing the line to support additional load.
