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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