Q178 / 180 · KCET 2010 · Physics

In a series resonant R-L-C circuit, the voltage across $R$ is $100 \mathrm{~V}$ and the value of $\mathrm{R}=1000 \Omega$. The capacitance of the capacitor is $2 \times 10^{-6} \mathrm{~F}$; angular frequency of AC is $200 \mathrm{rad} \mathrm{s}^{-1}$. Then the potential difference across the inductance coil is

  • A
    $100 \mathrm{~V}$
  • B
    $40 \mathrm{~V}$
  • Correct answer
    C
    $250 \mathrm{~V}$
  • D
    $400 \mathrm{~V}$

Solution

The current in the circuit $$ \begin{aligned} \mathrm{i} &=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \\ &=\frac{100}{1000}=0.1 \mathrm{~A} \end{aligned} $$ At resonance, $$ \begin{aligned} \mathrm{V}_{\mathrm{L}} &=\mathrm{V}_{\mathrm{C}}=\mathrm{iX}_{\mathrm{C}}=\frac{\mathrm{i}}{\omega \mathrm{C}} \\ &=\frac{0.1}{200 \times 2 \times 10^{-6}} \\ &=250 \mathrm{~V} \end{aligned} $$

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

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