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A supply of 400V is connected across capacitors... - JAMB Physics 2019 Question

A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge

A

8 x 10\(^{-4}\)C

B

4 x 10\(^{-2}\)C

C

8 x 10\(^{-3}\)C

D

4 x 10\(^{-8}\)C

correct option: a
C\(_T\) =C\(_1\) × C\(_2\)
C\(_1\) + C\(_2\)

 

=3 × 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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