A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
- halved
- same
- Zero
- double
Answer
(4) double
Solution
At resonance XL = XC ⇒ Z = R
I0 = V0R
R R2 (resonance unchanged, 0 = 1 LC) ⇒ I0' = V0R/2 = 2I0
Answer: (4)