The net capacitance of the circuit shown above is
Attached Figure / Diagram
A
8.0µF
B
6.0µF
C
4.0µF
D
2.0µF
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Capacitance in parallel.
\(C_4 = C_1 + C_2\) = 2 + 2 = 4µf.
\(\frac{1}{C_T} = \frac{ 1}{C_3} + \frac{1}{C_4}\)
= \(\frac{ 1}{4} + \frac{1}{4}\)
= \(\frac{ 1 + 1}{4}\)
\(\frac{1}{C_T} = \frac{2}{4}\)
\(C_T = \frac{4}{2}\) = 2µf.
Source: Myschool Past Questions Archive (Record #41699)