Class 12 Physics Alternating Current Power associated with Capacitor

Power associated with  Capacitor

  • Voltage and current associated with the  capacitor  are  given as V = Vm sinωt  and  I = Im sin(ωt +(∏/2)) respectively .
  •  Instantaneous power p = VI =  Vm Im sinωt sin(ωt +(∏/2))
  • Average Power PC =  (1/T) 0T∫ p dt
  • =  (1/T)  0T∫ Vm Im sinωt sin (ωt +(∏/2))  dt
  • After simplifying  the above expression:-
  • PC = 0. This implies the  average  power supplied to a capacitor over one complete cycle is zero.

Phasor  diagram  and   Graphical representation

Problem:- A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V,60 Hz supply. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum?

Answer:-

Capacitance of the capacitor, C = 100 μF = 100 × 10−6F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

(a) Frequency of oscillations, ν= 60 Hz

Angular frequency ω = 2πν= 2π x 60 rad/s,

For a RC circuit, we have the relation for impedance as:

Z = √(R2 +1/(ω2C2))

Peak voltage, V0 = V√2 = 110√2

Maximum current is given as:

I0= (V0/Z)

= V0/(√(R2 +1/(ω2C2))

=(110√2)/( √ (40)2 + (1/(120 π)2 x (10-4)2)

=(110√2)/( √(1600 +(10)8/(120 π)2)

=3.24A

(b) In a capacitor circuit, the voltage lags behind the current by a phase angle of Φ. This angle is given by the relation:

tan Φ = (1/ωC)/(R) = (1/ ωCR)

=(1)/( 120 π x 10-4 x 40) = 0.6635

Φ =  tan-1 (0.6635) = 33.560

=(33.560)/(180) rad

Therefore Time lag = (Φ/ω)

=(33.56 π)/(180 x 120 π)

=1.55 x 10-3 s

=1.55ms

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.

 

Problem:- A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the RMS value of the current in the circuit.

Answer:-

Capacitance of capacitor, C = 60 μF = 60 × 10−6 F

Supply voltage, V = 110 V

Frequency, ν = 60 Hz

Angular frequency, ω = 2πν

Capacitive reactance, XC = (1/ωC)  = 1/(2πνC)

=1/(2π × 60 × 60 × 10−6) Ω

RMS value of current is given as:

I =(V)/(XC)

= (220) /(2π × 60 × 60 × 10−6) = 2.49 A

Hence, the rms value of current is 2.49 A.

 

Problem:- A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (RMS and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

Answer:-

The capacitive reactance is

XC  = (1/ 2πνC )

=(1/2π (50Hz)(15.0x10-6F) )

=212 Ω

The RMS current is

I =(V/XC)

=(220V)/(212 Ω)

The peak current is

Im = √2 I = (1.41)(1.04A) = 1.47 A.

This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

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