A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
C\(_T\) = \(\frac{C_1 \times C_2}{C_1 + C_2}\)
= \(\frac{3 \times 6}{3 + 6}\)
= \(\frac{18}{9}\) = 2μf
Q = CV
⇒ 2 × 10\(^{-6}\) × 400
⇒ 800 × 10\(^{-6}\)C = 8 × 10\(^{-4}\)C
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