Question
Download Solution PDFTwo capacitance C1 = 150 ± 2.4 μF and C2 = 120 ± 1.5 μF. What is the limiting error of the resultant capacitance C?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 3): 3.9 μF
Concept:
when two capacitors C1, C2, connected in parallel.The resultant capacitance C is
C = C1 +C2 F
The limiting error of the resultant capacitance is the sum of individual capacitance limiting error
Calculation:
Given
C1 = 150 ± 2.4 μF
C2 = 120 ± 1.5 μF.
The Two capacitors are connected in parallel so the resultant capacitance
C = C1 + C2
= 150 ± 2.4 μF + 120 ± 1.5 μF.
= 270 ± 3.9 μF.
The limiting error of the resultant capacitance is the sum of individual capacitance limiting error
The limiting error = 3.9 μF.
Last updated on Jun 24, 2025
-> WBPSC JE recruitment 2025 notification will be released soon.
-> Candidates with a Diploma in the relevant engineering stream are eligible forJunior Engineer post.
-> Candidates appearing in the exam are advised to refer to the WBPSC JE syllabus and exam pattern for their preparations.
-> Practice WBPSC JE previous year question papers to check important topics and chapters asked in the exam.