Advertisements
Advertisements
Question
For a series LCR-circuit, the power loss at resonance is ______.
Options
I2 R
I2 ω C
`"V"^2/(omega "C")`
`"V"^2/(omega "L" - 1//omega"C")`
MCQ
Fill in the Blanks
Solution
For a series LCR-circuit, the power loss at resonance is I2 R.
Explanation:
At resonance, the impedance of the circuit is equal to the resistance. Therefore,
Pav = I2 R
shaalaa.com
Is there an error in this question or solution?
