A supply of 400V is connected across capacitors of 3\muf and 6\muf 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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
C\(_T\) = \(\frac{C_1 \times C_2}{C_1 + C_2}\)
= \(\frac{3 \times 6}{3 + 6}\)
= \(\frac{18}{9}\) = 2\muf
Q = CV
⇒ 2 \times 10\(^{-6}\) \times 400
⇒ 800 \times 10\(^{-6}\)C = 8 \times 10\(^{-4}\)C
Source: Myschool Past Questions Archive (Record #55090)