Question

A device 'X' is connected to an ac source V = V₀ sin ωt. The variation of voltage, current and power in one cycle is shown in the following graph:

(a) Identify the device 'X'.
(b) Which of the curves A, B and C represent the voltage , current and the power consumed in the circuit? Justify your answer.
(c) How does its impedance vary with frequency of the ac source? Show graphically.
(d) Obtain an expression for the current in the circuit and its phase relation with ac voltage.

Answer 1

(a) The device 'X' is a pure capacitor.

(b) Curve A represents power P=VI, where the amplitude is equivalent to the multiplication of amplitudes of V and I curve. Curve B is the sine curve and represents voltage and curve C is a cosine curve representing the current.
The full cycle of the graph consists of two positive and two negative symmetrical areas. So the average power consumed in the circuit is zero.

(c) The AC impedance of a capacitor is known as Reactance and as we are dealing with capacitor circuits, more commonly called Capacitive Reactance. The graph shows variation of a capacitive reactance with frequency.

 `z= x_c = 1/omega_c`

`x_c=1/(2pif_c)`

`v_c prop 1/f`

(d)

AC through capacitor:- Let us consider a capacitor with capacitance C be connected to an AC source with an emf having an instantaneous value

`E = E_0 sinomegat`..(i)

Due to this emf, a charge will be produced and it will charge the plates of a capacitor with a positive and negative charge. if potential difference  across the plates of a capacitor is V then

`V=q/C " or" q=CV`

The instantaneous value of current in the circuit

`I = (dq)/dt=(d(CE))/(dt)`  (∵ V = E)

`=d/(dt)(CE_0sinomegat)`    `(because E = E_0sinomegat)`

`=CE_0cosomegat xx omega`

`=E_0/(1/omega_c)sin(omegat)`   `[thereforecosomegat = sin(pi/2+omegat)]`

`I=E_0/(1/omega_c)sin(omegat+pi/2)`...............(ii)

`I` will be maximum when `sin(omegat+pi/2)=1` so that `I=I_0`

where peak value of current `I_0E_0/(1/omega_c)`

`I=Isin(omegat+pi/2)`..............(iii)

from (1) and (iii) it is clear current leads the voltage by a phase angle of `pi/2(a_0)`