A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
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Correct Answer: Option A
Explanation:
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
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