Question
Download Solution PDFWhat is Ceq for the given circuit ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
1, 5 and 2 μF capacitors are connecter in series.
\(\therefore \frac{1}{{C_{eq}}} = \frac{1}{1} + \frac{1}{5} + \frac{1}{2}\)
\( = \frac{{10 + 2 + 5}}{{10}} = \frac{{17}}{{10}}\)
\(\therefore Ceq = \frac{{10}}{{17}}μ F\)
12 μF and \(\)\(\frac{{10}}{{17}}\) μF capacitors are connected in parallel
\(∴ {C_{eq}} = 12 + \frac{{10}}{{17}} = \frac{{214}}{{17}} = 12.58μ F\)
12.58 μF, 0.4 μF and 0.8 μF are connected in series
\(∴ \frac{1}{{Ceq}} = \frac{1}{{12.58}} + \frac{1}{{0.4}} + \frac{1}{{0.8}}\)
∴ Leq = 0.26 μF
Parallel, 7 µF and 5μF are connected in series
\(∴ \frac{1}{{Leq}} = \frac{1}{7} + \frac{1}{5}\)
∴ Leq = 2.916 μF
∴ Ceq =3.176 MF
∴ Correct option is (2)
Last updated on May 28, 2025
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.