Question on: JAMB Physics - 2019
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
Ask EduPadi AI for a detailed answer
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
Add your answer
Please share this, thanks!
No responses