If a capacitor of capacitance C is connected across the alternating source, the spontaneous charge on the capacitor,
q = CVC = CV0 sin wt
and the instantaneous current i going through it, is shown by:
![1211_Alternating current.png](https://www.expertsmind.com/../CMSImages/1211_Alternating current.png)
or i = i0 sin (wt + p/2)
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![1432_Alternating current1.png](https://www.expertsmind.com/../CMSImages/1432_Alternating current1.png)
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Here, V = i/wC
This equation gives that the quantity is the effective ac resistance or the capacitive reactance of the capacitor and is represented as XC. It has unit as ohm. Thus, X =1/wC
It is seems that the current has the voltage by 90° or the potential drop across the capacitor lags the current passing it by 90°.
Diagram gives V and i as functions of time t.
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![1364_Alternating current2.png](https://www.expertsmind.com/../CMSImages/1364_Alternating current2.png)
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Inductor in an AC Circuit: Consider a pure inductor of self inductance L and zero resistance connected to an alternation source. Again we suppose that an instantaneous current i = i0 sin wt flows through the inductor. Although there is no stoppage, there is a potential difference VL in between the inductor terminals a and b because the current changes with time, giving rise to self induced emf.
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![254_Alternating current3.png](https://www.expertsmind.com/../CMSImages/254_Alternating current3.png)
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![614_Alternating current4.png](https://www.expertsmind.com/../CMSImages/614_Alternating current4.png)
Equation (iii) shows that effective ac resistance, i.e., productive reactance of inductor is,
XL = wL
And the maximum current, i=V/XL
The unit of XL is also ohm.
From Equations. (i) and (iii) we see that the voltage across the inductor leads the current passing through it by 90°.
Diagram defines VL and i as functions of time.
![1191_Alternating current5.png](https://www.expertsmind.com/CMSImages/1191_Alternating%20current5.png)
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