Q152 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Physics

A capacitor of capacitance $100 \mu \mathrm{F}$ and a coil of resistance $20 \Omega$ and inductance $12.5 \mathrm{mH}$ are connected in series with a $220 \mathrm{~V}, \frac{200}{\pi} \mathrm{Hz}$ AC source. The maximum value of instantaneous current in the circuit is

  • A
    $20 \mathrm{~A}$
  • B
    $10 \mathrm{~A}$
  • Correct answer
    C
    $11 \mathrm{~A}$
  • D
    $15 \mathrm{~A}$

Solution

$\mathrm{x}_{\mathrm{L}}=\omega \mathrm{L}$ $$ \begin{aligned} & =2 \pi \mathrm{fL}=2 \pi \times \frac{200}{\pi} \times 12.5 \times 10^{-3} \\ & =400 \times 12.5 \times 10^{-3}=5 \Omega \end{aligned} $$ $\begin{aligned} & \text { and, } \mathrm{x}_{\mathrm{C}}=\frac{1}{400 \times 100 \times 10^{-6}} \\ & =25 \Omega \\ & \mathrm{i}_0=\frac{\mathrm{V}_0}{\sqrt{\left(\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{C}}\right)^2+\mathrm{R}^2}}=\frac{220 \sqrt{2}}{\sqrt{(5-25)^2+20^2}} \\ & =\frac{220 \sqrt{2}}{20 \sqrt{2}}=11 \mathrm{~A}\end{aligned}$

Hence, the correct answer is option: 'C'.

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