Lesson AC Capacitance · RC circuits and power factor
RC parallel circuits
In an RC parallel circuit the voltage across R and C is the same, but branch currents differ in phase. Resistive current is in phase with voltage; capacitive current leads voltage by 90°. Total current is the vector sum of the branch currents. Impedance is lower than either branch alone when both conduct.
Same voltage, different currents
Parallel rule: ET is common. Each branch obeys its own limit:
IR = ET / R (in phase with ET) IC = ET / XC (leads ET by 90°)
Total line current:
IT = √(IR² + IC²)
Then the circuit impedance seen by the source:
Z = ET / IT
Z will be less than the smaller of R and XC when both branches take significant current — more paths, more total current, lower equivalent ohms.
Phase angle in parallel RC
cos θ = IR / IT tan θ = IC / IR
Current still leads voltage overall (leading PF), but the angle depends on how large IC is compared with IR. A big capacitor branch (small XC) pulls θ toward 90° lead.
Power in parallel RC
True power still comes only from the resistive branch:
P = ET × IR or P = IR² × R
Capacitive branch:
VARsC = ET × IC
Apparent power:
VA = ET × IT
Same three power ideas as series — different path to the numbers.
Why parallel shows up so often
- Line filters and EMI snubbers across supplies
- Capacitors across control coils (noise)
- PF capacitors across plant buses (with inductors/motors elsewhere)
- Bypass capacitors on electronics boards
Many “capacitor across the line” mental models are parallel RC (the R being the load).
Calculating values step by step
- Find XC = 1/(2πfC) if given C.
- Find IR and IC from ET.
- Find IT from the current triangle.
- Find Z = ET/IT.
- Find P, VARsC, VA, PF = P/VA = IR/IT.
Compare with series RC: there you started from Z and one current; here you start from one voltage and two currents.
Series vs parallel quick contrast
| Series RC | Parallel RC | |
|---|---|---|
| Common quantity | Current | Voltage |
| Triangle sides | ER, EC or R, XC | IR, IC |
| Z formula start | √(R²+XC²) | ET/IT after current sum |
| PF | R/Z | IR/IT |
Numbers you should be able to work cold
ET = 120 V, R = 40 Ω, XC = 30 Ω:
IR = 3 A IC = 4 A IT = 5 A Z = 24 Ω PF = IR/IT = 0.6 leading P = 360 W VA = 600 VA
Notice Z (24 Ω) is less than both R and XC — parallel paths always invite that outcome when both conduct.
When the capacitor branch opens
IC → 0, IT → IR, PF → 1, noise may rise. On a motor with a parallel run capacitor, an open can means the motor loses its designed phase shift — symptoms look like a bad capacitor, not a “better PF.”
When the capacitor branch shorts
The branch looks like a near short across the line — breaker or fuse operation. That is a failed dielectric, not normal IC.
Field case
Situation. A 120 V control transformer secondary feeds a relay coil (resistive/inductive, but approximate R for this story) with a suppressor capacitor paralleled across the coil. A tech removes the capacitor because “it draws extra current,” then the PLC input false-triggers from coil kickback noise.
What happened. The capacitor’s IC is real but small and intentional. Removing it raised noise and inductive kick issues.
Applied lesson. Parallel capacitive current is often by design. Measure IR vs IC; do not rip out suppression parts without understanding the branch.
In the field
Symptom
Line current higher than resistive load alone; noise returns after “cleanup”
Where to look
Caps across loads, filters, PF cans on same bus
Likely causes
- Normal IC
- shorted capacitor (IR-like fault current)
- open capacitor (noise/PF worse)
What to measure
- Branch currents if accessible
- ET
- compute expected IC = ET/XC
What not to do
- Assume any parallel capacitor is wasted energy without checking function
Checklist
- I use common ET for both branches
- I compute IR and IC separately
- I find IT = √(IR² + IC²)
- I find Z = ET/IT
- I compute PF = IR/IT (leading)
- I contrast this method with series RC