Capacitive reactance is the opposition a capacitor gives to alternating current. Its formula is XC = 1/(omega C) = 1/(2 pi f C), and its unit is the ohm, same as resistance. Memory hook: "C-a-p-Capacitor Cuts high frequency, so XC is small when f is big" (XC is inversely proportional to both f and C).
Capacitive reactance XC falls as frequency rises (a 1/f curve). At very low frequency or DC, XC is huge and blocks current; at high frequency XC is small and lets AC pass. Its unit is the ohm.
Your doubts, answered
Is capacitive reactance the same thing as resistance?
No. Both are measured in ohm and both limit current, but they are different. Resistance R does not depend on frequency and it turns electrical energy into heat. Capacitive reactance XC depends on frequency (XC = 1/(2 pi f C)) and it stores and returns energy without heating the capacitor. So XC opposes current but wastes no power.
Why does XC decrease when frequency increases?
Look at the formula: XC = 1/(2 pi f C). Frequency f is in the denominator. When f goes up, the whole fraction becomes smaller, so XC goes down. Physically, at high frequency the AC reverses so fast that the capacitor barely charges up before the voltage flips, so it opposes the current less. That is why a capacitor allows high-frequency AC to pass easily.
What is the unit of capacitive reactance?
The unit is the ohm (the same as resistance and inductive reactance). You can check this: 1/(farad x per-second) gives ohm. In NEET numericals, always convert C into farad (microfarad x 10^-6) and use omega = 2 pi f in rad/s to get XC in ohm.
Does a capacitor have reactance in a DC circuit?
For steady DC the frequency f = 0, so XC = 1/(2 pi x 0 x C) = infinity. That means a fully charged capacitor blocks steady DC completely (acts like an open switch). This is the classic result: a capacitor blocks DC but allows AC.
Does XC depend on the applied voltage or current?
No. XC depends only on the frequency f (or omega) and the capacitance C. It does NOT depend on the voltage or current values. Changing the source voltage changes the current, but XC stays the same as long as f and C are fixed.
⚠️ The NEET trap ✗ Students write XC = omega C or XC = 2 pi f C, copying the inductive reactance form XL = omega L. ✓ Capacitive reactance is the INVERSE: XC = 1/(omega C) = 1/(2 pi f C). The 1-over is the whole point. XC gets SMALLER at high frequency, while XL gets larger. 🧠 Capacitor Cuts, so XC is a Cutdown = 1 OVER omega C. Inductor Lifts, so XL = omega L (no fraction).
Real NEET questions
NEET 2020
A 40 microfarad capacitor is connected to a 200 V, 50 Hz AC supply. The rms value of the current in the circuit is, nearly:
A · 2.5 A ✓
B · 25.1 A
C · 1.7 A
D · 2.05 A
Solution: Step 1: Find omega. omega = 2 pi f = 2 pi x 50 = 100 pi rad/s.
Step 2: Find capacitive reactance. XC = 1/(omega C) = 1/(100 pi x 40 x 10^-6) = 1/(4 pi x 10^-3) = about 79.6 ohm.
Step 3: Find rms current. I_rms = V_rms / XC = 200 / 79.6 = about 2.51 A, which is nearly 2.5 A. Answer: A.
NEET 2024
A 10 microfarad capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (pi = 3.14):
A · 0.93 A ✓
B · 1.20 A
C · 0.35 A
D · 0.58 A
Solution: Step 1: Find capacitive reactance. XC = 1/(omega C) = 1/(2 pi x 50 x 10 x 10^-6) = 1000/pi ohm (about 318 ohm).
Step 2: The source gives V_rms = 210 V, so peak voltage V0 = sqrt2 x V_rms.
Step 3: Peak current i0 = V0 / XC = (sqrt2 x 210) / (1000/pi) = (1.414 x 210 x 3.14)/1000 = about 0.93 A. Answer: A.
Solved Alternating Current NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
