RC Circuits: Resistance, Reactance, and Power Factor Analysis

The primary focus of this study is the relationship between resistance ( $R$ ) and reactance ( $X_C$ ), specifically how these components interact to produce the phase angle ( $\theta$ ) of a circuit.
The phase angle ( $\theta$ ) is a trigonometric value derived from the circuit's capacitor and resistor values. It describes how much the reactive current (capacitive or inductive) is offset from the voltage.
In a series circuit, there is only one path for current, which dictates that the current ( $I$ ) must be the same throughout every component.

Because resistance and reactance are $90$ degrees out of phase, their values cannot be added algebraically. Instead, a vector relationship is used, often visualized as a triangle.
Impedance ( $Z$ ), representing the total opposition to current, is found using the Pythagorean theorem:
- $Z = \sqrt{R^2 + X_C^2}$
This relationship applies to voltage as well. Since capacitors oppose changes in voltage, there is a phase shift, and voltage drops across a resistor and a capacitor do not add up linearly to the total supply voltage.

True Power ( $P$ ): This is the actual power consumed by the load, measured in Watts ( $W$ ). It corresponds to the resistive part of the circuit.
Apparent Power ( $S$ ): Also referred to as theoretical power, it is measured in Volt-Amps ( $VA$ ). It is the product of the supply voltage and total current ( $V \times I$ ).
Reactive Power ( $Q$ ): Measured in Volt-Amps Reactive ( $VAR$ ). This power represents the energy stored and released by the reactive components (capacitors or inductors).
These three types of power form the Power Triangle, where apparent power is the hypotenuse, true power is the adjacent side, and reactive power is the opposite side.

Consider an RC circuit with the following known values:
- Resistor ( $R$ ) = $12\,\Omega$
- Capacitor Reactance ( $X_C$ ) = $16\,\Omega$
- Applied Voltage ( $V$ ) = Unknown at start, but used to find current.
Calculating Total Impedance ( $Z$ ):
- $Z = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\,\Omega$
Calculating Phase Angle ( $\theta$ ):
- Using trigonometry (SOH CAH TOA), if we use the ratio of the opposite ( $X_C$ ) to the hypotenuse ( $Z$ ):
- $\sin(\theta) = \frac{16}{20} = 0.8$
- $\theta = \sin^{-1}(0.8) \approx 53^\circ$
Ohm's Law Applications:
- If the current ( $I$ ) is noted as $12\,A$ throughout the series circuit:
- Voltage across resistor ( $V_R$ ) = $12\,A \times 12\,\Omega = 144\,V$
- Voltage across capacitor ( $V_C$ ) = $12\,A \times 16\,\Omega = 192\,V$
- Total supply voltage ( $V_{total}$ ) = $12\,A \times 20\,\Omega = 240\,V$
Power Calculations:
- Reactive Power ( $VARs$ ) = $192\,V \times 12\,A = 2304\,VAR$
- True Power ( $Watts$ ) = $144\,V \times 12\,A = 1728\,W$
- Apparent Power ( $VA$ ) = $240\,V \times 12\,A = 2880\,VA$ (Note: transcript also mentions $2080\,VA$ and $2.8\,kW$ during discussion).

To find the specific capacitance ( $C$ ) in Farads ( $F$ ), the capacitive reactance formula is rearranged:
- $X_C = \frac{1}{2\pi f C}$ becomes $C = \frac{1}{2\pi f X_C}$
Assuming a standard frequency ( $f$ ) of $60\,Hz$ and $X_C = 16\,\Omega$ :
- $C = \frac{1}{2 \times \pi \times 60 \times 16} \approx 0.0001658\,F$
Unit Conversion:
- To convert Farads to Microfarads ( $\mu F$ ), multiply by $1,000,000$ ( $10^6$ ).
- $0.0001658 \times 10^6 \approx 165.8\,\mu F$ (Transcript uses the rounded value of $169\,\mu F$ ).

Power Factor ( $PF$ ): This is the ratio of true power to apparent power, representing the efficiency of the circuit.
- $PF = \frac{\text{True Power}}{\text{Apparent Power}}$
Using the example values:
- $PF = \frac{1728\,W}{2880\,VA} = 0.6$
Expressed as a percentage, the power factor is $60\%$ . This means $60\%$ of the supplied power is doing useful work, while $40\%$ is being wasted or "dissipated away" as reactive power.
Power Factor Correction: Most industrial loads are inductive. To improve efficiency and bring the phase angle closer to zero (where true power equals apparent power), capacitors are used to offset the inductive reactance ( $X_L$ ), as $X_C$ and $X_L$ act in opposite directions (180 degrees apart, both 90 degrees from resistance).

When capacitors are connected in series:
- The total capacitance is the inverse of the sum of the inverses ( $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \dots$ ).
- However, their reactive power values ( $VARs$ ) in a circuit add up directly because they are in the same line.
Example with Multiple Capacitors:
- If True Power = $5\,W$ and Apparent Power is calculated as $10.62\,VA$ from two capacitors in series:
- $PF = \frac{5}{10.62} \approx 0.47$ or $47\%$
- $\theta = \cos^{-1}(0.47) \approx 61.95^\circ$

Q: How would I find the current through the resistor?
- A: You use Ohm's Law ( $V=IR$ ). If you have the total resistance (impedance $Z$ ) and the supply voltage, you can find the total current, which is the same as the current through the resistor in a series circuit.
Q: Does a purely capacitive load differ from an inductive load?
- A: Yes, primarily in the direction of the phase shift. While both are $90$ degrees out of phase with resistance, they point in opposite directions on the vector diagram, allowing them to cancel each other out.
Q: Is there a way to get apparent power without other values?
- A: Apparent power is generally a calculated vector sum of true and reactive power ( $S = \sqrt{P^2 + Q^2}$ ) or measured as the total product of plant supply voltage and total current. It is the power that the utility company must supply and the customer must pay for.
Administrative Reminder: Exercise E is due at the beginning of the next class. It covers purely capacitive and RC circuits.