A 5 µF capacitor is connected in series with a 10 µF

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 Multiple Choice QuestionsMultiple Choice Questions

271.

Three capacitors 3 µF, 6 µF and 6 µF are connected in series to a source of 120 V. The potential difference, in volt, across the 3 µF capacitor will be

  • 24

  • 30

  • 40

  • 60


272.

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

  • the energy stored in the capacitor decreases

  • the capacitance of the capacitor increases

  • the charge on the capacitor decreases

  • the voltage across the capacitor increases


273.

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant K. The remaining half contains air as shown in the figure. The capacitor is now given a charge Q. Then

  • electric field in the dielectric-filled region is higher than that in the air-filled region.

  • on the two halves of the bottom plate the charge densities are unequal.

  • charge on the half of the top plate above the air-filled pant is QK + 1

  • capacitance of the capacitor shown above is 1 + K C02, where C0 is the capacitance of the same capacitor with the dielectric removed.


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274.

A 5 µF capacitor is connected in series with a 10 µF capacitor. When a 300 V potential difference is applied across this combination, the total energy stored in the capacitors is

  • 15 J

  • 1.5 J

  • 0.15 J

  • 0.10 J


C.

0.15 J

According to question, the figure can be drawn as below

     

The equivalent capacitance,

      1Ceq = 1C1 + 1C2 = 15 + 110 = 2 + 110 = 310  Ceq  = 103 μF

Now, the energy stored in the capacitor is

U = 12 CV2 = 12 × 103 × 10-6 × 300 × 300   = 310 × 2 = 320    = 0.15 J


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