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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current5 Marks
Question
Define the term capacitive reactance. Show graphically the variation of capacitive reactance with frequency of applied alternating voltage. An ac voltage $\text{V = V}_0\sin\omega\text{t}$ is applied across a pure capacitor of capacitance C. Find an expression for current flowing through it. Show mathematically the current flowing through it leads the applied voltage by angle $\frac{\pi}{2}.$
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Answer


Capacitive Reactance: The opposition offered by a capacitor alone to the flow of alternating current through it is called the capacitive reactance.
It is denoted by $X_C$ Its value is $\text{X}_{\text{C}}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$
The graph of variation of capacitive reactance with frequency is shown in figure.
Phase Difference between Current and Applied voltage in Purely Capacitive Circuit:

Circuit Containing Pure Capacitance: Consider a capacitor of capacitance C; connected to an alternating voltage source as shown.
As ac voltage changes in magnitude and direction periodically with a definite frequency; therefore the plates of capacitor get charged, discharged and charged in opposite direction, discharged continuously (Fig. b). Therefore the flow of alternating current in the circuit is maintained. The instantaneous voltage,
$\text{V = V}_0\sin\omega\text{t} \ ...(\text{i})$
Let q be the charge on capacitor and i, the current in the circuit at any instant, then instantaneous potential difference,
$\text{V}=\frac{\text{q}}{\text{c}} \ ...(\text{ii})$
$\text{q}=\text{CV}_0\sin\omega\text{t}$

The instantaneous current,
$\text{i}=\frac{\text{dq}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{CV}_{\text{0}}\sin\omega\text{t})=\text{CV}_{0}\frac{\text{d}}{\text{dt}}(\sin\omega\text{t})$
$=\text{CV}_{0}\omega\cos\omega\text{t}$
$\text{i}=\frac{\text{V}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}\cos\omega\text{t}=\frac{\text{V}_0}{\frac{1}{\omega\text{C}}}\sin\Big(\omega\text{t}+\frac{\pi}{2}\Big)$
$\text{i}=\text{I}_0\sin\Big(\omega\text{t}+\frac{\pi}{2}\Big) \ ...(\text{iii})$
where $\text{i}_0=\frac{\text{V}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$ = Peak value of A.C. ...(iv)
Also comparing (i) and (iii), we note that the current leads the applied emf by an angle $\frac{\pi}{2}$ This is shown graphically in fig. (c).

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