Test: AC Applied Across Capacitor - JEE MCQ

We can use a capacitor of suitable capacitance as a choke coil because the average power consumed per cycle in an ideal capacitor is zero. Therefore, like a choke coil, a condenser can reduce AC without power dissipation.

equation of alternating voltage , E = 200√2sin(100t)
compare this equation with E₀sin(ωt)
so,  E₀=200√2 volts
ω= 100 rad/s
now, E_rms=E₀/√2
= 200√2/√2 = 200 volts .
given, capacitance of capacitor, C = 10^-6 F
so, reactance of capacitor,  X_C=1/ω_C
=1/(10^-6x100)
=10⁴
so, the reading of Ammeter, I=E_rms/X_C
= 200/10⁴ A = 20 × 10^-3 A = 20mA
hence, option (b) is correct.

The opposition offered by a capacitor for the flow of A.C is called capacitive reactance.
Xc = 1/wC
it's SI unit is ohm

As we know,
Average power in ac circuit is given by P=V_rms​i_rms​cosϕ 
For pure capacitive circuit ϕ=90^o so P=0

Natural Frequency of an LC Oscillator

- The natural frequency of an LC oscillator is determined by the values of the inductance (L) and capacitance (C) in the circuit.

- The LC oscillator is a type of oscillator circuit that uses a combination of inductors and capacitors to generate oscillations at a specific frequency.

- The natural frequency of an LC oscillator can be calculated using the formula:

f = 1 / (2 * π * √(LC))
 

P_av=(V₀i₀/2) cosθ
In ideal capacitor θ= -π2
Cosθ=Cosθ(-π/2)=0
Therefore
P_av=0

Capacitive reactance (symbol XC) is a measure of a capacitor’s opposition to AC (alternating current). Like resistance it is measured in ohms, but reactance is more complex than resistance because its value depends on the frequency (f) of the signal passing through the capacitor. Reactance is also inversely proportional to the value of capacitance (C), i.e. the value of XC at any frequency will be less in larger capacitors than in smaller ones. All capacitors have infinitely high values of reactance at 0Hz, but in large capacitors, the reactance falls to a low level at much lower frequencies than in smaller capacitors. Hence, larger capacitors are preferred in low frequency applications.

Impedance(X_C)=1/ω_C
=1/ 2πfC
=1/2πx50x10x10^-6
=10⁶/1000π
X_C=10³/π=(1000/π) Ω

Voltage lags current by 90° in a pure capacitive circuit. In a pure capacitive circuit, the instantaneous power may be positive or negative. As with the simple inductor circuit, the 90-degree phase shift between voltage and current results in a power wave that alternates equally between positive and negative.

Xc ≈ 0.026 Ohms