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 answer is 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