Using the diagram above, calculate the effective capacitance of the circuit
-
1.56μf
- 3.00μF
- 3.71μF
- 9.00μF
The correct answer is: A
Explanation
\(C_1\) = 3μF + 4μF = 7μF ( capacitance in parallel )
\(\frac{1}{ C} = \frac{ 1}{2} + \frac{1}{7} = \frac{ 7 + 2 }{14} = \frac{9}{14}\)
∴ C = \(\frac{14}{9}\) = 1.56μF