Question

An alternating emf is applied across a capacitor. Show mathematically that current in it leads the applied emf by a phase angle of $π$ /2. What is its capacitive reactance ? Draw a graph showing the variation of capacitive reactance with the frequency of the a.c. source.

Solution

Let an alternating emf, E = E₀ sin ωt is applied across a capacitor of capacitance C.
The current flowing in the circuit transfers charge to the plates of the capacitor due to which a potential difference develops across its plates. Also, assume that 'q' be the charge on each plate of the capacitor at any instant t.

Therefore, potential difference across the plates of capacitor,

V = \frac{q}{C} .

At every instant potential difference V must be equal to the applied emf.

i.e.,

V = \frac{q}{C} = E = E_{0} \sin ωt

...(i)

[∵ V = E = E_{0} \sin ωt]

q = {CE}_{0} \sin ωt

∴

Instantaneous current,

I = \frac{dq}{dt} = \frac{d}{dt} ({CE}_{0} \sin ωt) = {CE}_{0} \cos ωt . ω

I = \frac{E_{0}}{\frac{1}{ωC}} \sin (ωt + \frac{π}{2})

...(ii)

[∵ \sin (θ + \frac{π}{2}) = \cos θ]

The current is maximum, i.e.,

I = I_{0}

when

\sin (ωt + \frac{π}{2}) = 1

From equation (i),

I_{0} = \frac{E_{0}}{\frac{1}{ωC}} \times 1 = \frac{E_{0}}{\frac{1}{ωC}}

...(iii)

Putting in equation (ii), we get

I = I_{0} \sin (ωt + \frac{π}{2})

Comparing equations (i) and (iii), we see that in an a.c. circuit containing capacitor only, current leads the emf by a phase angle of

π / 2

.

Comparing (iii) with I₀ = E₀/R, we find 1/ωC is the effective resistance of the capacitive a.c. circuit. It is called capacitive reactance.

i.e.,

χ_{C} = \frac{1}{ωC} = \frac{1}{2 πfC} .

Fig. Variation of capacitive reactance with frequency.

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An alternating emf is applied across a capacitor. Show mathematically that current in it leads the applied emf by a phase angle of $π$ /2. What is its capacitive reactance ? Draw a graph showing the variation of capacitive reactance with the frequency of the a.c. source.