1.

A 120V, 60 HzA.C. is connected across a non - inductive resistace of 400 Omega an un-known capacitor joined in series. The voltage across resistance is 66.3 V. The voltage drop across capacitor is

Answer»

120 V
66.3 V
53.7 V
100 V

Solution :`V^(2)=V_(C )^(2)+V_(R )^(2)rArr (120)^(2)=(66.3)^(2)+V_(C^(2))`
`therefore V_(C )=100` VOLT


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