JAMB - Physics (1982 - No. 39)

what is the resultant capacity of the circuit
1.5µF
18.0µF
6.0µF
6.8µF
8.83µF
Explanation
1μf and 5μf are parallel;
Ceff = 1 + 5 = 6μf
6μf, 4μf and 4μf are in series.
total capacitance of the circuit in series( \(\frac{1}{C_T}\)) = \(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\)
= \(\frac{1}{4} + \frac{1}{4} + \frac{1}{6}\)
= \(\frac{ 6 + 6 + 4}{ 24}\) = \(\frac{ 16}{ 24}\)
\(\frac{1}{C_T}\) = \(\frac{ 16}{ 24}\)
\(C_T\) = \(\frac{24}{ 16}\) = 1.5μf.
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