StreakPeaked· Practice

ExamsJEE AdvancedPhysics

In an AC circuit, a resistor R is connected in series with an inductor L = 0.5 H and a capacitor C = 200 microfarad. The source voltage is V = 200 sin(60t) volts. The voltage amplitude across the capacitor (Vc) shows a maximum value as the source frequency is varied. If the source frequency at which this maximum occurs equals the given source frequency of 60 rad/s, then the value of R is expressed as 40*sqrt(x) ohms. Find x.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 2

Solution

The capacitor voltage in a series RLC is maximised at omega_Cmax = sqrt(1/LC - R²/(2L²)). Setting this equal to the driving frequency 60 rad/s and using omega0 = 100 rad/s gives R² = 2L²(omega0² - 60²) = 2*(0.25)*(10000-3600) = 3200, so R = 40*sqrt(2) ohms, giving x = 2.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →