Correct Answer: Option D

Explanation

C = \(\frac{\text{Q}}{\text{V}}\)  => Q = CV : ∴ V = \(\frac{\text{Q}}{\text{C}}\)
∴V\(_1\)  = \(\frac{\text{Q}}{C_1}\); V\(_2\) = \(\frac{\text{Q}}{C_2}\);
V\(_3\) = \(\frac{\text{Q}}{C_3}\)
for series
\(\frac{1}{\text{C}}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) + \(\frac{1}{C_3}\)
= \(\frac{1}{2} + \frac{1}{6} + \frac{1}{3}\)

= \(\frac{3 +1 + 2}{6}\)

= \(\frac{6}{6}\) = 1

∴ C = 1μF = 1.0 x 10\(^{-6}\)F
∴ Q = CV = 1.0 x 10\(^{-6}\) x 12
= 12.0 x 10\(^{-6}\)C
∴V\(_1\) = \(\frac{\text{Q}}{C_1}\)

= \(\frac{12 \times 10^{-6}}{2 \times 10^{-6}}\)

  = 6V

V\(_2\) = \(\frac{\text{Q}}{C_2}\)

= \(\frac{12 \times 10^{-6}}{6 \times 10^{-6}}\)

  = 2V
V\(_3\) = \(\frac{\text{Q}}{C_3}\)

= \(\frac{12 \times 10^{-6}}{3 \times 10^{-6}}\)

= 4V

∴ 6V, 2V and 4V

There is an explanation video available below.

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