A series LCR circuit is connected to a variable frequency 230 V source. L = 5H, C = 80 µF & R = $40 \Omega$. The potential drop across R at resonating frequency is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 230 V
At resonance, the inductive resistance $(X_L)$ and capacitive resistance $(X_C)$ are equal:
$X_L=X_C$
$⇒2πfL=\frac{1}{2πfC}$
$⇒f=\frac{1}{2π\sqrt{LC}}=\frac{1}{2π\sqrt{5×60×10^{-6}}}≃12.57Hz$
At resonance, the impedance of the circuit $Z=R$ is -
$I=\frac{V_{supply}}{Z}=\frac{V_{supply}}{R}$ [By ohm's law]
$=\frac{230V}{40Ω}=5.75A$
$∴V_R=I.R$
$=(5.75×40)$
$=230V$