Q64 / 84 · JEE Advanced 2010 (Paper 1) · Physics

An AC voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased

  • A
    the bulb glows dimmer
  • Correct answer
    B
    the bulb glows brighter
  • C
    total impedance of the circuit is unchanged
  • D
    total impedance of the circuit increases

Solution

$Z=\sqrt{R^2+X_C^2}: I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}: P=I_{\mathrm{rms}}^2 R$ where, $\quad X_C=\frac{1}{\omega C}$ As $\omega$ is increased, $X_C$ will decrease or $Z$ will decrease. Hence $I_{\mathrm{mms}}$ or $P$ will increase. Therefore, bulb glows brighter. Hence the correct option is (b).

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

Relevant Questions