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TS EAMCET · Physics · Current Electricity

A capacitor of capacitance 4 μF is charged to a potential difference of 6 V with a battery. The battery is then removed and in its place another capacitor of capacitance 8 μF is introduced and the circuit is closed. The potential difference attained by each of the capacitors in V is

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(A) 2

Step-by-step Solution

Detailed explanation

When two capacitors are connected in series then their equivalent capacitance is, Ceq=C1C2C1+C2=4×84+8=83 μF Q=CeqV=83×6=16 μC Now V1=QC1=164=4 V & V2=QC2=168=2 V ∴  ∆V=4-2=2 V
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