Question

A series LCR circuit has
$L = 0.01H, R = 10Ω\ and\ C = 1µF$
and it is connected to ac voltage of amplitude (Vm)50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately

• A. 466 mA
• B. 312 mA
• C. 238 mA
• D. 196 mA

Answer 1

Answer: (C)

238 mA

Answer 2

The correct answer is (C) : 238 mA
ω = 0.4 ω0....(i)
$⇒ I = \frac{V}{Z} = \frac{50}{\sqrt{R^2+(ωL-\frac{1}{ωC})^2}}...(ii)$
⇒ I = 238 mA