JAMB - Physics (2024 - No. 109)
Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.
0.0009μF
0.0002μF
0.0005μF
0.0001μF
Explanation
Combined capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\)
The two capacitors have capacitance of 0.0003μF and 0.0006μF 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\(_T\) = 0.0002μF.
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