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Those same two capacitors in series carry 12 V across the pair. The voltage across the 3 \( \mu \)F capacitor is

  • A.4 V
  • B.6 V
  • C.12 V
  • D.8 V

Answer: D

The same charge sits on both, so the voltages divide in inverse proportion to the capacitances and the smaller capacitor takes the larger share, two thirds of twelve volts. This is the opposite of how a series resistive divider behaves with respect to resistance.

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