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


Question

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

Options

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

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

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

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

The correct answer is A.

Explanation:

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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Dicussion (1)

  • 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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