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A 120 V, 60 Hz power source is connected...

A 120 V, 60 Hz power source is connected across an `800 Omega` non inductive resistance and an unkown capacitance in series. The voltage drop across the resistor is 102 V.
(a) What is the voltahe drop acorss the capacitor ?
(b) What is the reactance of the capacitor ?

Text Solution

Verified by Experts

Here, `V = 120 V, v =60 Hz`,
`R = 800 Omega C = ?, V_(R ) = 102 V, V_(C ) = ? X_(C ) = ?`
As `v_(R ) = I R`
`:. I = (V_(R ))/(R ) = (102)/(800) = 0.128 A`
As `V = sqrt(V_(R )^(2) + V_(C )^(2))`
`:. V_(C) = sqrt(V^(2) - V_(R )^(2)) = sqrt(120^(2) - 102^(2))`
`= 63.2 V`
As `V_(C ) = I X_(C )`
`:. X_(C ) = (V_(C ))/(I) = (63.2)/(0.128) = 493.8 Omega`
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