Question
Download Solution PDFThe equivalent capacitance between points A and B in the below-shown figure will be ________ μF.
Answer (Detailed Solution Below) 6
Detailed Solution
Download Solution PDFCONCEPT:
- The capacitance of series - The capacitance series are in series we have;
\(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} +....\)
Here we have Cs are equivalent is series, C1, C2, C3,.... is the capacitance in series in 1, 2,3... capacitor.
- The capacitance of parallel - The capacitance series are in series we have;
Cp = C1 +C2 +C3 +...
Here we have CP are equivalent is series, C1, C2, C3,.... is the capacitance in parallel in 1, 2,3...capacitor.
CALCULATION:
Note: The two capacitors connected with the same wire without junction will vanish from the circuit. Therefore the revised circuit is;
As we can see in the above image that the three capacitances are parallel to each other than according to the parallel capacitance we have;
Cp = C1 + C2 + C3
⇒ Cp = 8 + 8 + 8
⇒ Cp = 24 μF
In the above figure capacitance, Cp, and 8 μF are in series with each other.
∴ \(\frac{1}{C_s} =\frac{1}{C_p}+\frac{1}{8}\)
⇒ \(\frac{1}{C_s} =\frac{1}{24}+\frac{1}{8}\)
⇒ \(\frac{1}{C_s} =\frac{1+3}{24}\)
⇒ \(\frac{1}{C_s} =\frac{4}{24}\)
⇒ Cs = 6 μF
Hence, Cs = 6 μF
Last updated on May 23, 2025
-> JEE Main 2025 results for Paper-2 (B.Arch./ B.Planning) were made public on May 23, 2025.
-> Keep a printout of JEE Main Application Form 2025 handy for future use to check the result and document verification for admission.
-> JEE Main is a national-level engineering entrance examination conducted for 10+2 students seeking courses B.Tech, B.E, and B. Arch/B. Planning courses.
-> JEE Mains marks are used to get into IITs, NITs, CFTIs, and other engineering institutions.
-> All the candidates can check the JEE Main Previous Year Question Papers, to score well in the JEE Main Exam 2025.