The resultant capacitance in the fig above is
Attached Figure / Diagram
A
15.0µF
B
9.8µF
C
1.3µF
D
0.8µF
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Capacitors in parallel = 4\(\mu\)f + 2 \(\mu\) f = 6\(\mu\) f
Capacitors in series =\(\frac{1}{C_{eff}}\) = \(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}\)
\(\frac{1}{C_{eff}}\) = \(\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{4}\) = \(\frac{6 + 4 + 2 + 3}{12}\)
\(\frac{1}{C_{eff}}\) = \(\frac{15}{12}\)
C\(_{eff}\) = \(\frac{12}{15}\) = 0.8\(\mu\) f
Source: WAEC / JAMB / NECO Classroom Archive (Record #41732)