Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.
A
0.0009μF
B
0.0002μF
C
0.0005μF
D
0.0001μF
Verified Correct Answer: B.
Physical Principles & Numerical Solution
Combined capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\)
The two capacitors have capacitance of 0.0003\muF and 0.0006\muF respectively
\(\frac{1}{C_T}\) = \(\frac{1}{0.0003}\) + \(\frac{1}{0.0006}\) = \(\frac{0.0002 + 0.0001}{0.0006}\)
\(\frac{1}{C_T}\) = \(\frac{0.0003}{0.0006}\)
C\(_T\) = 0.0002\muF.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70682)