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In series LCR circuit voltage drop acros...

In series LCR circuit voltage drop across resistance is 8V, across inductor is 6V and across capacitor is 12V. Then

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.

Text Solution

Verified by Experts

The correct Answer is:
D

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.
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