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Alternating Current — Physics STD 12 Science — Question

Rajasthan 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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