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:

$230 V$

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]

$=\frac{230V}{40Ω}=5.75A$

$∴V_R=I.R$

$=(5.75×40)$

$=230V$