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Two capacitors C1 (6µF) and C2 (12µF) are in series across a 180 volts D.C supply. Calculate the charges on C1 and C2 respectively.

Correct answer: D. 720 x 10^-6 C, 720 x 10^-6 C

  • A. 120 x 10^6 C, -420 x 10^6 C
  • B. 320 x 10^-6 C, 420 x 10^-6 C
  • C. 420 x 10^-6 C, 320 x 10^-6 C
  • D. 720 x 10^-6 C, 720 x 10^-6 C
  • E. 620 x 10^-6 C, 420 x 10^-6 C

Explanation

Charge remains same in series and charges of both capacitors are only same in Option D and so it is the right answer.You can even calculate the charge using this;net capacitance in series:1/C= 1/C1 + 1/C21/C = 1/12 + 1/6 C= 4x10-6 FC=Q/V4x10-6 = Q/180Q= 720 x 10-6 C

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About Electrostatics

Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.

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