Q63 / 90 · KVPY 2017 (19 Nov SB/SX) · Physics

Consider the R-L-C circuit given below. The circuit is driven by a $50 \mathrm{~Hz}$ AC source with peak voltage $220 \mathrm{~V} .$ If $\mathrm{R}=400 \Omega, \mathrm{C}=200 \mu \mathrm{F}$ and $\mathrm{L}=6 \mathrm{H}$, the maximum current in the circuit is closest to Figure

  • Correct answer
    A
    $0.120 \mathrm{~A}$
  • B
    $0.55 \mathrm{~A}$
  • C
    $1.2 \mathrm{~A}$
  • D
    $5.5 \mathrm{~A}$

Solution

$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times 6=600 \pi \Omega$ $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi 80 \times 200 \times 10^{-6}}=\frac{100}{2 \pi}=\frac{50}{\pi} \Omega$ $\mathrm{I}_{\max }=\frac{\mathrm{V}}{\mathrm{z}}=\frac{220}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}$ $=\frac{220}{\sqrt{400^{2}+\left(600 \pi-\frac{50}{\pi}\right)^{2}}}$ $\mathrm{I}_{\max }=0.120 \mathrm{~A}$

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

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