A120 v, 60 hz power source is connected in series across an 800 ω resistance and an unknown capacitance in series. the voltage drop across the resistor is 102 v. find (a) the voltage drop across the capacitor; and (b) the reactance of the capacitor.

a)Vc= 63.21 V

b)Xc= 495.756  Ω

Explanation:

Assume that voltage across the resistance is Vr and capacitor Vc

So

Now by putting the values

Vc= 63.21 V

We know that for resistance

Vr= I R

102 = I x 800

I=0.1275 A

Let reactance of capacitor is Xc

So

Vc= Xc . I

63.21 = 0.1275 x Xc

Xc= 495.756  Ω

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