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

battery with hi vtge high val resist in series

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form