Capacitive Circuitsalternating Currents Topic

JEE Advanced

  1. Capacitor
  • Capacitive reactance: XC=12πfC
  • Current flowing: I=VXC

XC=12π×50 Hz×100×106 F=318.31Ω

I=200 V318.31Ω=0.628 A

  1. AC Circuit with Inductor and Capacitor:
  • Impedance: Z=R2+(XLXC)2
  • Current flowing: I=VZ

XL=2πfL=2π×50 Hz×0.1 H=31.42Ω

Z=102+(31.42318.31)2=300.85Ω

I=200 V300.85Ω=0.665 A

  1. Step-up Transformer:
  • Voltage across secondary: Vs=NsNpVp

Vs=200  turns100 turns×100 V=200 V


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  1. Capacitor
  • Capacitive reactance: XC=12πfC

  • Current flowing: I=VXC XC=12π×50 Hz×100×106 F=318.31Ω

I=200 V318.31Ω=0.628 A

  1. AC Circuit with Inductor and Capacitor
  • Impedance: Z=R2+(XLXC)2

  • Current flowing: I=VZ

XL=2πfL=2π×50 Hz×0.1 H=31.42Ω

Z=102+(31.42318.31)2=300.85Ω

I=200 V300.85Ω=0.665 A

  1. Step-up Transformer:
  • Voltage across secondary: Vs=NsNpVp

Vs=200 turns100 turns×100 V=200 V

  1. AC Generator:
  • Peak voltage: Vpeak=Vrms2 Vpeak=220 V×2=311 V

  • Peak current: Ipeak=VpeakR Ipeak=311 V100Ω=3.11 A

  1. Inductor with AC Source:
  • Irms=VrmsZ,whereZ=R2+XL2

XL=2πfL=2π(60 Hz)(0.1 H)=37.7Ω

Z=102+37.72=38.6Ω

  • Current amplitude: Irms=120 V38.6Ω=3.11 A
  1. Capacitor with AC Source:

With XC=12πfC=26.52Ω

  • Current amplitude: I=Vrms1XC=120 V26.52Ω=4.52 A


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