WAEC - Physics (1999 - No. 32)
Using the diagram above, calculate the effective capacitance of the circuit
1.56μf
3.00μF
3.71μF
9.00μF
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
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