In series L-C-R circuit voltage drop across resistance is 8 V and across inductor is 6 V and across capacitor is 12 V . Then

Options

A. voltage of the source will be leading current in the circuit
B. voltage drop across each element will be less than the applied voltage.
C. power factor of circuit will be `4//3`
D. none of these.

Answer

D

Explanation

Since `cos phi =(R )/(Z) =(IR)/(IZ) =8/10=4/5`
(also `cos phi` can never be greater than 1)
Hence, (c ) is wrong .
Also, `IX_(C ) gtIX_(L) implies (X_C) gt(X_L)`
`:.` Current will be leading
In and LCR circuit,
`V=sqrt((v_(L)-v_(C ))^(2)+c_(R )^(2))=sqrt((6-12)^(2)+8^(2))`
`V=10,` which is less than voltage drop across capacitor.