Question
Download Solution PDFदिए गए नेटवर्क में यदि R = 1Ω है, तो ओममीटर मान निर्धारित कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
जब R = 1 Ω, 4 Ω इसके साथ श्रेणी में हो।
4 Ω + 1 Ω = 5 Ω
अब, 5 Ω || 20 Ω
\(\frac{20 \times 5}{20+5} = 4 ~\Omega\)
अंत में 2 Ω,12 Ω 4 Ω के साथ श्रेणी में होगा,
इसलिए, 4 + 2 + 12 =18 Ω
R = 1Ω होने पर दिए गए नेटवर्क में ओममीटर मान 18 Ω होता है।
Last updated on May 9, 2025
-> PGCIL Diploma Trainee result 2025 will be released in the third week of May.
-> The PGCIL Diploma Trainee Answer key 2025 has been released on 12th April. Candidates can raise objection from 12 April to 14 April 2025.
-> The PGCIL DT Exam was conducted on 11 April 2025.
-> Candidates had applied online from 21st October 2024 to 19th November 2024.
-> A total of 666 vacancies have been released.
-> Candidates between 18 -27 years of age, with a diploma in the concerned stream are eligible.
-> Attempt PGCIL Diploma Trainee Previous Year Papers for good preparation.